TG 


UC-NRLF 


ESI    SEE 


MAXIMUM 


AME      BRIDGES, 


THE 


VAN  NOSTRAND 


No.  ±2.  THEORY  OF  VOUSSOIR  ARCHES.   By 

Prof.  Wm.  Cain.     Third  edition,  revised  and  enlarged. 


THE   VAN   NOSTRAND   SCIENCE    SERIES 

No.  13.     GASES  MET  WITH  IN  COAL  MINES. 

By  J.  J.  Atkinson.  Third  edition,  revised  and  enlarged, 
to  which  is  added  The  Action  of  Coal  Dusts  by  Edward 
H.  Williams,  Jr. 

No.  14.    FRICTION  OF  AIR  IN  MINES.      By  J.  J. 

Atkinson.     Second  American  edition. 

No.  15.    SKEW  ARCHES.     By  Prof.  E.  W.  Hyde, 

C.E.     Illustrated.     Second  edition. 

No.  16.    GRAPHIC    METHOD     FOR     SOLVING 

Certain  Questions  in  Arithmetic  or  Algebra.  By  Prof. 
G.  L.  Vose.  Third  edition. 

*No.  17.    WATER     AND     WATER-SUPPLY.      By 

Prof.  W.  H.  Corfield,  of  the  University  College,  London. 
Second  American  edition. 

No.  18.    SEWERAGE     AND     SEWAGE     PURIFI- 

cation.  By  M.  N.  Baker,  Associate  Editor  "Engineer- 
ing News."  Fourth  edition,  revised  and  enlarged. 

No.  19.     STRENGTH       OF       BEAMS       UNDER 

Transverse  Loads.  By  Prof.  W.  Allan,  author  of 
"Theory  of  Arches."  Second  edition,  revised. 

No.  20.    BRIDGE  AND  TUNNEL  CENTRES.     By 

John  B.  McMaster,  C.E.     Second  edition. 

No.  21.    SAFETY    VALVES.     By  Richard  H.  Buel, 

C.E.     Third  edition. 

No.  22.    HIGH  MASONRY  DAMS.      By  E.  Sher- 
man Gould,  M.  Am.  Soc.  C.E.     Second  edition. 

No.  23.    THE    FATIGUE    OF    METALS    UNDER 

Repeated  Strains.  With  various  Tables  of  Results  and 
Experiments.  From  the  German  of  Prof.  Ludwig 
Spangenburg,  with  a  Preface  by  S.  H.  Shreve,  A.M. 

No.  24.  A  PRACTICAL  TREATISE  ON  THE 

Teeth  of  Wheels.  By  Prof.  S.  W.  Robinson.  Third 
edition,  revised,  with  additions. 

No.  25.     THEORY      AND      CALCULATION      OF 

Cantilever  Bridges.     By  R.  M.  Wilcox. 

No.  26.  PRACTICAL  TREATISE  ON  THE  PROP- 

erties  of  Continuous  Bridges.     By  Charles  Bender,  C.E. 

No.  27.    BOILER  INCRUSTATION  AND  CORRO- 

sion.  By  F.  J.  Rowan.  New  edition.  Revised  and 
partly  rewritten  by  F.  E.  Idell. 

*No.  28.   TRANSMISSION  OF  POWER  BY  WIRE 

Ropes.     By  Albert  W.  Stahl,  U.S.N.     Fourth  edition, 
revised. 
No.  29.     STEAM  INJECTORS;  THEIR  THEORY 

and  Use.  Translated  from  the  French  by  M.  Leon 
Pochet. 


THE   VAN  NOSTRAND   SCIENCE  S  ERIES 

No.  30.    MAGNETISM  OF  IRON  VESSELS  AND 

Terrestrial  Magnetism.     By  Prof.  Fairman  Rogers. 

No.  31.    THE  SANITARY  CONDITION  OF  CITY 

and  Country  Dwelling-houses.  By  George  E.  Waring, 
Jr.  Third  edition,  revised. 

No.  32.     CABLE-MAKING     FOR     SUSPENSION 

Bridges.     B.  W.  Hildenbrand,  C.E. 

No.  33.    MECHANICS     OF    VENTILATION.      By 

George  W.  Rafter,  C.E.     Second  edition,  revised. 

No.  34.      FOUNDATIONS.          By       Prof.       Jules 

Gaudard,  C.E.  Translated  from  the  French.  Second 
edition. 

No.  35.   THE      ANEROID      BAROMETER;      ITS 

Construction  and  Use.  Compiled  by  George  W. 
Plympton.  Eleventh  edition,  revised  and  enlarged. 

No.  36.    MATTER     AND  MOTION.     By  J.  Clerk 

Maxwell,  M.A.     Second  American  edition. 

*No.  37.    GEOGRAPHICAL     SURVEYING;      ITS 

Uses,  Methods,  and  Results.  By  Frank  De  Yeaux 
Carpenter,  C.E.  * 

No.  38.    MAXIMUM     STRESSES     IN     FRAMED 

Bridges.  By  Prof.  William  Cain,  A.M.,  C.E.  New 
and  revised  edition. 

No.  39.    A     HANDBOOK     OF    THE     ELECTRO- 

Magnetic  Telegraph.  By  A.  E.  Loring.  Fourth  edi- 
tion, revised. 

*No.  40.       TRANSMISSION     OF     POWER     BY 

Compressed  Air.      By  Robert  Zahner,  M.E. 

No.  41.      STRENGTH        OF       MATERIALS.      By 

William  Kent,  C.E.,  Assoc.  Editor  "Engineering  News.'' 
Second  edition. 

No.  4?.     THEORY        OF        STEEL  -  CONCRETE 

Arches,  and  of  Vaulted  Structures.  By  Prof.  Wm. 
Cain.  Fifth  edition,  thoroughly  revised. 

No.  43.    WAVE     AND     VORTEX     MOTION.      By 

Dr.  Thomas  Craig,  of  Johns  Hopkins  University. 

No.  44.     TURBINE       WHEELS.      By  Prof.   W.   P. 

Trowbridge,  Columbia  College.  Second  edition.  Re- 
vised. 

No.  45.     THERMODYNAMICS.      By    Prof.    C.    F. 

Hirshfeld.     Second  edition,  revised  and  corrected. 

No.  46.     ICE-MAKING      MACHINES.      From  the 

French  of  M.  Le  Doux.  Revised  by  Prof.  J.  E.  Denton, 
D.  S.  Jacobus,  and  A.  Riesenberger.  Sixth  edition, 
revised. 


MAXIMUM  STRESSES 


IX 


FRAMED   BRIDGES, 


BY 


0-A.X3ST, 

MEMBEB    AM.     SOC.    C.  E., 

PROFESSOR  OF  MATHEMATICS  IN  THE  UNIVERSITY  OF 
NORTH  CAROLINA. 


NEW  YORK: 
D.  VAN  NOSTRAND  COMPANY, 

25  PARK  PLACE 
1914 


Copyright,  1897,  by  D.  VAN  NOSTRAND  Co. 


PREFACE. 

THE  first  edition  of  this  work  was 
published  in  Van  Nostrand's  Magazine  for 
1878,  and  was  largely  concerned  with  the 
comparison  of  the  weights  of  bridges  and 
their  most  economical  depths.  These  sub- 
jects have  now  been  practically  solved  by 
bridge  engineers,  and  the  result  has  been 
the  elimination  of  many  types  of  bridge 
truss  once  popular  and  the  retention,  by 
the  principle  of  the  "survival  of  the 
fittest,"  of  certain  leading  forms  that  have 
proved  most  economical  and  otherwise  de- 
sirable. 

From  these  considerations,  it  was  thought 
best  to  confine  the  present  edition  to  the 
discussion  of  those  types  most  used  at 
present,  and  to  leave  out  any  comparison  of 
weights  and  extended  discussions  as  to  min- 
imum material.  The  work  has  therefore 
been  entirely  rewritten  upon  a  new  basis, 
the  aim  being  to  prepare  an  elementary 
treatise  on  the  maximum  stresses  in  bridge- 


359973 


11.  PKEFACE. 

trusses  of  selected  types,  both  for  uniform 
and  wheel  loads,  which  might  serve  either 
as  an  independent  short  treatise  on  the  sub- 
ject or  as  an  introduction  to  the  larger  trea- 
tises. The  subject-matter  has  been  given 
in  more  detail  perhaps  than  usual,  for  ex- 
perience in  teaching  has  shown  that  in  no 
other  way  can  a  student  quickly  and  surely 
master  its  first  principles.  Attention  is  es- 
pecially called  to  the  exact  treatment  for 
wheel  loads.  The  aim  throughout  has  been 
to  aid  the  student  by  presenting  the  subject 
in  a  simple,  clear  and,  at  the  same  time, 
thorough  manner. 

CHAPEL  HILL,  N.  C., 

April,  1897. 


TABLE  OF  CONTENTS. 

ARTICLES.  PAGE. 

1 .     D  efinitions,  Weights  of  Bridges 7 

7.     Laws  of  Mechanics,  Maximum  Shears. .  18 
14.     Stresses  in  a  Warren  Truss  with  Sus- 
penders  28 

19.     Chord   Stresses,   Method   of  Moments.  38 

21.     Method  of  Chord  Increments 43 

25.     Formulas  for  Chord  Stresses 49 

28.     Warren  Truss  (deck  and  through) 57 

32.     Shear  Formulas,  approximate  and  exact.  65 

38.  Through  Bridge,  Pratt  Truss 73 

39.  Through  Bridge,  Howe  Truss 81 

40.  Deck  Bridge,  Pratt  Truss 86 

41.  Deck  Bridge,  Howe  Truss 92 

43.  Whipple  Truss 94 

44.  Lattice  Truss 103 

45.  Wheel  Loads,  Reactions  and  Moments. .  110 

50.  Graphical  Method 116 

51.  Position    of   Load,    giving    maximum 

moment 121 

53.     Application  to  Pratt  Truss 130 

55.  Position  of  Live  Load,  giving  maximum 

shears  in  a  beam  and  a  truss 137 

56.  Application  to  Pratt  Truss 142 

58.    Hip  Vertical,  Floor  Beam  and  Stringer.  149 

60.     Warren  Girder 156 

62.     Stresses  due  to  Wind  Pressure 161 

68.     Theoretical  Section  Areas 174 

APPENDIX. 

Most  economical  height  of  trusses  hav- 
ing parallel  chords 177 


MAXIMUM  STRESSES 

IN 

FBAMED  BEIDGES. 


1.  A  simple  truss  is  a  structure  com 
posed  of  straight  members,  connected  at 
their  ends  by  pins  or  other  means,  and  de- 
signed to  transfer  loads  on  it  to  the  abut- 
ments at  either  end,  by  which  it  is  sup- 
ported, by  the  resistances  of  the  members 
to  longitudinal  tension  or  compression. 

The  ideal  truss  has  the  members  so  con- 
nected at  their  joints  that  the  resultant  of 
the  stress  on  any  cross-section  of  a  mem- 
ber will  pass  through  its  centre  of  gravity. 
This  ideal  is  impossible  to  attain,  except 
for  vertical  members,  as  for  all  others  the 
weight  of  the  member  will  cause  non-uni- 
form stress  on  the  cross-section.  If  the 
lines  through  the  centres  of  gravity  of  the 
cross- sections,  or  axes  of  members,  do  not 
meet  in  a  point  at  the  joint  where  the 


a- e rabers  assemble,  or  if  the  resultant 
stress  on  a  member  is  not  axial  because  it 
is  connected  to  other  members  only  along 
a  part  of  its  section,  or  if  the  members  are 
not  free  to  rotate  at  a  joint  under  loads, 
additional  stresses  are  caused  over  those 
borne  by  the  members  of  the  ideal  truss. 
Such  additional  stresses  are  called  second- 
ary, and  should  always  be  considered  in 
the  final  design. 

In  this  little  treatise,  space  will  not  per- 
mit dealing  with  other  than  the  ideal 
truss. 

2.  An  important  requisite  of  a  well-de- 
signed truss  is,  that  the  geometric  figure 
formed  by  the  axes  of  the  members  should 
be  of  invariable  form.  If  we  call  m  the 
number  of  bars  or  members  necessary  to 
this  end,  and  n  the  number  of  joints  or 
apices,  then  for  a  plane  figure,  assume  one 
side  as  fixed  in  position.  From  the  two 
apices,  at  its  ends,  we  can  fix  another  apex 
with  two  new  sides  ;  then  another,  with 
two  new  sides,  from  any  two  apices  pre- 
viously fixed,  and  so  on ;  therefore,  to  each 
of  the  (n — 2)  apices  other  than  the  first 


two,  corresponds  two  sides,  so  that  the  total 
number  of  sides,  m,  just  necessary  to  fix 
the  apices,  is  given  by  the  following  rela- 
tion: 

m  =  2  (n  —  2)  +  1  =  2  n  —  3  (1) 

If  the  number  of  sides  exceeds  (2  n — 3) 
the  extra  number  are  called  "  superfluous." 
A  less  number  will  give  a  figure  that  can 
change  its  shape  without  changing  the 
length  of  its  sides. 


As  an  application,  note  that  in  the 
Warren  Truss  (fig.  1),*  m  =  23,  n  —  13,  SD 
that  eq.  (1)  obtains,  and  the  figure  has 
enough  sides  to  strictly  define  the  form 
and  no  more. 

In  the  Queen  Post  Truss  without  cross- 

*  The  axes  of  the  members  only  are  shown  in  this  and 
other  figures  of  trusses. 


§3]  10 

bracing  in  the  middle  panel  (fig.  2)  m  —  8 
n  =  6  and  eq.  (1)  is  not  satisfied,  one  side  be- 
ing lacking.  The  stiffness  of  the  chord  here 


is  relied  on  to  prevent  deformation,  so  that 
the  truss  is  an  imperfect  one  from  one 
standpoint,  but  it  is  very  efficient  for  a  roof 
truss  or  a  highway  bridge  in  practice. 

3.  A  Framed  Bridge  is  generally  com- 
posed of  two  or  more  trusses,  as  A  d  B 
(fig.  1),  which  lie  in  vertical  planes,  parallel 
to  the  line  of  road,  the  trusses  being  con- 
nected together  by  transverse  bracing,  as 
well  as  by  the  floor  beams,  which  extend 
from  an  apex  of  one  truss  to  a  correspond- 
ing apex  on  the  other,  on  which  stringers, 
placed  parallel  to  the  line  of  road,  rest, 
which  support  the  cross- ties  and  rails  of 
a  railroad  bridge,  or  the  flooring  of  a 
highway  bridge. 

The  upper  and  lower  members  of  a  truss 


11  [§3 

are  called  the  upper  and  lower  chords,  the 
bracing  between  them  the  web. 

The  upper  -chord  is  always  in  com- 
pression, the  lower  chord  in  tension,  but 
the  web  members  are  generally  succes- 
sively struts  and  ties.  Those  web  mem- 
bers acting  always  as  ties,  are  sometimes 
called  main  ties\  those  acting  always  as 
struts,  main  braces  or  posts;  while  those 
acting  under  one  distribution  of  the  live 
load  as  struts,  and  under  another  as  ties, 
are  called  counter -br  aces  \  the  term  counter 
now  being  generally  restricted  to  those 
web  members  that  are  not  strained  appre- 
ciably by  the  dead  load,  but  are  strained 
when  the  live  load  is  distributed  in  a 
certain  manner. 

If  the  roadway  is  supported  by  the  top 
chords  (of  two  trusses,  say)  the  bridge  is 
called  a  deck  bridge ;  if  by  the  lower  chords 
it  is  called  a  through  bridge,  if  there  is 
overhead  bracing  between  the  top  chords; 
otherwise  it  is  termed  a  pony  truss. 

The  trussing,  in  vertical  planes,  connect- 
ing opposite  apices  of  the  two  trusses  of  a 
deck  bridge  is  called  sway  bracing.  When 


§4]  12 

a  through  bridge  is  more  than  25  feet 
high,  the  overhead  bracing  between  trusses 
in  vertical  planes  is  also  called  sway  brac- 
ing. The  lateral  bracing  lies  in  horizon- 
tal planes,  and  connects  the  two  top 
chords  together,  also  the  two  lower  chords. 
The  part  of  a  bridge  between  two  adja- 
cent joints  or  apices  is  called  a  panel,  a 
panel  length  being  the  distance  from  apex 
to  apex  corresponding.  In  this  book, 
bridges  with  two  trusses  will  alone  be  ex- 
amined. 

4.  A  truss  (fig.  1)  rests  upon  abutments 
at  A  and  B,  and  is  unsupported  the  dis- 
tance or  span,  A  B.  The  distance,  A  B,  is 
the  distance  between  end  pins,  centre  to 
centre. 

The  pressures  exerted  by  the  truss  on 
the  abutments  are  resisted  by  their  reac- 
tions, which  will  be  assumed  vertical,  as 
rollers  are  always  placed  under  one  end  of 
a  truss  over  75  feet  span.  Bow,  in  his 
"  Economics  of  Construction,"  has  given 
many  illustrations  of  the  effects  of  inclined 
reactions,  due  to  friction  at  the  abutments 
resisting  expansion  or  contraction  of 


§5]  13 

chords,  whether  caused  by  heat  or  the  pass- 
age of  loads. 

THE  WEIGHT  OF  BRIDGES. 
5.  The  weight  of  the  bridge  or  dead  load 
comprises  the  weight  of  the  two  trusses 
with  all  transverse  bracing,  floor  beams, 
stringers  and  flooring  system,  including  the 
planking  of  a  highway  bridge,  or  the  cross- 
ties,  rails,  guards,  spikes,  etc.,  of  a  railway 
bridge.  Merriman  gives  the  following 
formula  for  the  dead  load  per  linear  foot 
of  highway  bridges: 

^  =  140  +  125  +  0.2  bl—QAl (2) 

in  which  b  =  width  of  bridge  in  feet  (in- 
cluding sidewalks,  if  any),  I  =  span  in 
feet,  and  w  =  dead  load  in  pounds  per  lin- 
ear foot. 

The  formula  gives  weights  agreeing 
fairly  well  with  bridges  belonging  to  class 
*A  of  Waddel's  Specifications,  which  allows 
100  pounds  per  square  foot  of  flooring  for 
the  live  load  on  spans  from  0  to  100  feet, 
and  65  pounds  for  spans  of  350  feet  and 
over,  proportionate  amounts  being  allowed 
for  intermediate  spans,  provided  that  in  no 


14  [§5 

case  shall  the  live  load  per  linear  foot  be 
less  than  1800  pounds.  In  this  "  Class 
A,"  intended  for  city  bridges,  a  road 
roller,  having  12,000  pounds  on  the  front 
wheel,  4  feet  wide,  and  9000  pounds  on 
each  rear  wheel,  1  foot  8  inches  wide,  is  to 
be  used  in  computing  stresses  when  this 
loading  gives  the  maximum  stress  on  mem- 
bers, such  as  floor  beams,  stringers,  etc. 
The  distance  between  axles  of  front  and 
rear  wheels  is  11  feet,  and  the  distance 
between  the  central  planes  of  the  rear 
wheels  is  5  feet.  Johnson,  in  "Modern 
Framed  Structures,"  p.  44,  gives  a  diagram 
from  which  values  of  w  can  be  read  off 
more  accurately  than  by  formula  (2),  not 
only  for  case  A,  to  which  the  formula  ap- 
plies, but  also  for  cases  B  and  C,  which 
include  county  bridges. 

The  last-named  authority  is  quoted  be- 
low, for  the  weights  of  railroad  bridges 
corresponding  to  Cooper's  class,  "  Extra 
Heavy  A"  loading.  The  formulas  below, 
for  dead  load  per  linear  foot,  are  for  two 
trusses  only  and  single-track  bridges;  for 
double-track,  add  90  per  cent. 


§5]  15 

For  deck-plate  girders : 

w  =  91  +520 (3) 

For  lattice  girders : 

w  =  7  Z-f  600 (4) 

For  through  pin-connected  bridges : 

w  =  o  1+  750 (5) 

For  Howe  trusses  of  wood : 

w  =  6.21+  675 (6) 

In  all  these  formulas,  the  track  has  been 
included,  being  estimated  at  400  pounds 
per  linear  foot,  as  is  usual  in  computing 
stresses.  Its  average  value  is  nearer  300 
pounds. 

Example  I. — What  is  the  weight  of  a  through 
pin-connected  railroad  bridge  entire  (track  includ- 
ed) of  100  ft.  span  ? 

From  (5)  w  =  5  X  100  +  750  =  1250  pounds 
per  linear  foot ;  weight  of  bridge  is  1250  X  100  = 
125,000  pounds,  half  of  which,  or  62,500  Ibs.,  is 
carried  by  one  truss. 

Example  II. — What  is  the  weight  per  foot  of  a 
highway  bridge  of  100  ft.  span,  roadway  18ft.  wide? 
In  (2)  put  b  =  18,  1  =  100  and  get  w  =  676  .  %  , 
weight  on  one  truss  =  676  X  100  -s-  2  =  33,800 
pounds. 

6.  The  weight  acting  on  one  truss  is 
carried  to  the  apices  of  upper  and  lower 


§6]  16 

chords.  Thus,  in  fig.  1,  half  the  weight  of 
web  members,  the  weight  of  upper  chord 
and  half  the  upper  transverse  bracing  will 
be  supposed  borne  at  the  upper  chord 
apices.  Similarly  the  weights  of  half  th^ 
web  members,  lower  chord  and  lower 
transverse  bracing,  will  be  borne  at  lower 
chord,  and  in  addition  a  panel  length  of  the 
half  roadway.  Thus  at  an  apex  a,  if  b  and  c 
are  at  the  middle  of  adjoining  panels,  the 
weight  of  roadway  on  a  panel  length,  b  a  c, 
is  carried  first  to  the  floor  beam,  and  then 
half  this  weight  is  transferred  by  the  floor 
beam  to  apex  a,  and  the  other  half  to  the 
apex  of  the  truss  opposite. 

The  chord  that  carries  the  roadway  is 
thus  the  most  heavily  loaded ;  and  it  is  usual 
to  consider  that  it  carries,  at  its  apices,  two- 
thirds  of  the  dead  load  per  panel,  and  the 
other  chord  apices  one-third  of  the  same. 

If  we  suppose  fig.  1  to  represent  a  high- 
way bridge  of  100  span,  18  feet  between 
trusses,  we  have  just  found  (see  Ex.  II. 
above)  the  weight  per  linear  foot  of  bridge 
on  one  truss  to  be  338  pounds.  As  there 
are  6  panels,  the  weight  per  panel  is  338 


17  [§tf 

X  100  -T-  6  =  5633  pounds.  Two-thirds  of 
this,  or  3755  pounds,  is  supposed,  to  be  car- 
ried at  each  lower  apex,  and  one-third,  or 
1878  pounds,  at  each  apex,of  the  top  chord, 
except  at  the  ends,  where  only  half  or  939 
pounds  is  allowed,  as  the  end  apices  of  top 
chord  carry  only  half  the  weight  of  top  chord 
and  transverse  bracing  borne  at  the  other 
apices.  This  is  only  roughly  approximate, 
as  these  end  apices  really  carry  as  much  of 
the  web  as  the  interior  apices ;  still  the  re- 
sults are  near  enough  for  computation. 

If  we  call  the  dead  loads  sustained  at 
the  lower  apices  wl9  w%y  t03,  io4  and  w5,  each 
will  equal  3755  pounds.  If  we  call  the 
loads  sustained  at  the  apices  of  top  chord 
?/:6,  w7,  w8,  wQ9  wlo  and  t/Jn,  then  w6  = 
w^  •=.  939,  and  w7  =  w6  —  w9  =  wlQ  =  1878 
pounds.  The  half-panel  loads  in  end  lower 
panels  are  borne  directly  to  the  abutments 
at  A  and  B,  and  are  not  considered  in  the 
computation  of  stresses  in  the  truss  mem- 
bers. Thus  the  load  on  this  truss  will  be 
taken  as  5  X  3755  +  4  X  1878  +  2  X  939, 
or  5  X  5633  =  28,165  pounds.  The  reaction 
at  each  abutment  is  one-half  this,  or  14,082 


§7]  18 

pounds,  since  the  loads  are  symmetrically 
distributed  as  to  the  centre  of  the  truss. 

LAWS  OF  MECHANICS  AND  APPLICATIONS. 

7.  In  addition  to  the  principle  of  the 
triangle  of  forces,  the  following  laws  of 
mechanics  will  be  used: 

If  any  number  of  forces,  acting  on  a 
rigid  body  and  in  the  same  plane,  are  in 
equilibrium, — 

LAW  I.  The  algebraic  sum  of  their  mo- 
ments about  any  point  in  the  plane  of  the 
forces  is  zero. 

LAW  II.  The  algebraic  sum  of  their 
vertical  components  is  zero. 

LAW  III.  The  algebraic  sum  of  their 
horizontal  components  is  zero. 

Calling,  as  above,  w^  w^  . . .  ,  the  dead 
panel  loads  on  a  truss,  fig.  1,  and  xl9  x^  . . . , 
the  distances  to  the  right  abutment  (centre 
of  end  pin) ;  also  calling  ?//,  w"  .  . . ,  the 
part  of  any  live  load  borne  by  one  truss 
and  x'y  x\  .  .  . ,  the  distances  to  right 
abutment  from  their  centres  of  gravity  re- 
spectively, we  have  by  Law  I.,  calling  R 


19  [§7 

the  left  vertical  reaction,  Rt  the  reaction  at 
B,  and  I  the  span,  and  taking  B  as  a  centre 
of  moments : 

R  X  I  —  (wl  xl  -f-  w^  x9  4- . . .)  —  (wr  xf  + 
w"  x"  +  ...)  +  Rl  X  <?=0 

Whence, 

K  =  Cwi  gi  +  ^3  x*  +  •••)  +  M' g'  +  M" a"  +  --e(7) 

Or,  in  words :  to  find  the  left  vertical  re- 
action, add  the  products  obtained  by  mul- 
tiplying each  weight  or  load  by  its 
horizontal  distance  to  the  right  abutment, 
and  divide  the  sum  by  the  span. 

As  the  reaction  due  to  dead  load  on  one 
truss,  when  it  is  symmetrical  as  to  the 
centre,  is  always  one-half  the  sum  of  the 
dead  panel  loads,  it  is  generally  computed 
separately  and  added  into  the  reaction 
found  by  the  above  rule  for  the  live  load. 

In  railroad  bridges,  w  w"  . . . ,  are 
actual  loads  on  successive  wheels,  until  the 
specified  uniform  carload  is  reached,  when 
the  car-load  on  the  bridge  is  treated  as  a 
single  force  acting  through  its  centre  of 
gravity.  This  subject  will  be  developed 


§8]  20 

more  fully  later  on.  In  highway  bridges 
it  is  customary  to  compute  live  panel 
loads  for  the  load  per  square  foot  allowed 
on  the  clear  width  of  floor  between  trusses, 
and  to  regard  these  loads  per  panel  con- 
centrated at  the  apices  of  the  loaded 
chord.  This  may  be  called  the  method  of 
apex  loads,  which  last  are  supposed  to 
move  a  panel  length  at  a  time.  The  dis- 
tances x',  x,"  . . .  ,  as  well  as  I,  should  then 
be  given  in  panel  lengths,  regarding  a 
panel  length  as  the  unit  of  length. 

Ex.  When  the  above  formula  or  rule  is 
applied  to  the  dead  load  only  of  the  truss, 
why  does  it  give  R  -=.  half  dead  load  ? 

8.  If  we  call  W  the  total  load,  live  and 
dead,  on  the  truss,  we  have  by  Law  II. 

R  4-  RI  —  W  =  O (8) 

Therefore,  when  R  is  known  from  (7), 
Ra  can  be  found. 

9.  The   truss,  fig.  1,  will  now  be  sup- 
posed  to   be    severed  along  the  line  d  e. 
Suppose  the  part  of  the  truss  to  the  right 
of  d  e  removed,  and  that  forces  C,  D,  T, 
are   applied   at  the  cut  parts,    equal  and 


21  [§9 

directly  opposed  to  the  resistance  of  those 
members  (fig.  3).    These  forces  will  act  in 


Fig.5. 


the  direction  of  the  cut  members,  and  will 
exactly  represent  the  action  of  the  right 
part  of  the  truss  upon  the  left  at  the  sec- 
tion. 

The  forces  C,  D,  T,  the  reaction  R  and 
the  loads  (live  and  dead)  on  the  part  of 
the  truss  left  of  the  section  d  6,  thus  con- 
stitute a  system  of  forces  in  equilibrium 
acting  on  the  structure  A  d  e. 

Denote  the  vertical  component  of  D  by 
V,  its  horizontal  component  by  H;  then 
since  the  upper  chord  is  in  compression, 
the  lower  in  tension,  C  must  act  to  the 
left  and  T  to  the  right,  and  we  have,  for 
horizontal  chords,  by  Law  III.,  if  D  is  sup- 


§  10  ]  22 

posed  to  act  downwards,  and  .'.  H  to  the 

left, 

_C  —  H  +  T  =  O (9) 

If  D  acts  upwards,  H  is  directed  to  the 
right  and  is  plus  in  eq.  (9).  The  direction 
in  which  D  acts  is  shown  in  the  next 
article. 

10.  Call  the  sum  of  the  loads  on  the 
left  of  the  section  d  e,  2  w ;  then  the  differ- 
ence between  the  left  reaction  and  2  io9  or 

R—  2  w, 

is  called  the  shear  left  of  the  section 
Law  II.  applied  here  may  be  stated  as 
follows:  for  equilibrium,  if  the  shear 
(R  —  2  w)  is  plus,  it  indicates  that  it  acts 
up,  hence  V  must  act  down,  therefore  D 
acts  down,  and  the  diagonal  cut  by  the 
section  d  e^  is  in  compression ;  but  if  the 
section  is  taken  far  enough  to  the  right,  so 
that  (R —  2  w)  is  minus,  the  shear  just  to 
left  of  section  acts  down,  therefore  Y  must, 
for  equilibrium,  act  ^cp\  hence  D  acts  up 
and  the  diagonal  is  in  tension. 

The  above  laws  hold  when  the  top  of 
the  diagonal  leans  away  from  the  left 
abutment,  as  in  fig.  3. 


23  [§11 

11.  Let  the  section  be  now  passed  cut. 
ting  a  diagonal  whose  top  leans  toward^ 
the  left  abutment^  as  in  fig.  4. 


\ 


Fig.4 

Here,  if  the  shear  just  left  of  the  section 
(R  —  2  w}  is  plus,  it  acts  up,  .  *  .  Y  and 
therefore  D,  acts  down;  hence,  the  diag- 
onal cut  is  in  tension ;  but  if  (R  —  2  w)  is 
minus,  the  shear  acts  down,  and  hence  V 
and  D  act  up,  and  the  diagonal  cut  is  in 
compression. 

These  most  important  rules  should  be 
thoroughly  understood  by  the  reader,  who 
should  be  capable  of  immediately  applying 
them  to  find  the  amount  and  character  of 
the  stress  in  any  web  member.  The  rules 
apply  equally  in  case  any  of  the  web 
members  are  vertical.  In  this  case,  V  =  D 
=  stress  in  vertical  member. 


§  12]  24 

12.  If  i  is  the  inclination  of  any  diag- 
onal to  the  vertical,  the  stress  in  it  is, 

D  =  V  sec.  i\ 

hence,  since  V  is  equal  numerically  to  the 
shear  just  to  the  left  of  the  section  cutting 
the  diagonal,  we  have  the  important  rule: 

The  stress  in  any  diagonal  is  equal  to 
the  shear  on  it  multiplied  by  the  secant  of 
its  inclination  to  the  vertical. 

When  the  top  of  the  diagonal  leans 
towards  the  left  abutment ,  and  the  shear 
from  the  left  is  plus,  it  is  in  tension;  if  the 
shear  is  minus,  it  is  in  compression.  But 
if  the  top  of  the  diagonal  leans  from  the 
left  abutment  and  the  shear  is  plus,  the 
diagonal  is  in  compression;  if  the  shear  is 
minus,  it  is  in  tension. 

Example  I. — Draw  fig.  (1)  to  a  larger  scale  and 
write  the  panel  dead  loads  given  in  Art.  6  at  each 
apex,  then  number  the  diagonals  from  left  to 
right,  1,2,3...,  and  find  the  shears  and  character 
of  the  stresses  in  each  diagonal. 

In  the  table  below,  the  shear  in  (1)  is  the  left 
reaction,  14,082.  From  this  we  subtract  939  (load 
at  first  upper  apex)  to  find  shear  in  (2).  From 


[§13 


shear  in  (2)  we  subtract  load   3755  at  first  lower 
apex  to  find  shear  =  (B  —  ^  M)  in  (3),  and  so  on: 


DIAGONAL,. 

SHEAR. 

DIAGONAL. 

SHEAR. 

1 

+  14,082 

12 

-  14,083 

2 

+  13,143 

11 

-13,144 

3 

+    9,388 

10 

-    9,389 

4 

+    7,510 

9 

-    7,511 

5 

+    3,755 

8 

—    3,756 

6 

4-    1,877 

7 

-    1,878 

As  we  subtract  apex  loads  in  turn  from  the  pre- 
ceding shear,  we  find  the  shear  first  minus  in  (7). 
From  shear  in  (7)  subtract  1878  to  get  shear  in 
(8)  ;  from  this  3755  to  find  shear  in  (9),  and  so 
on  for  minus  shears.  The  character  of  the  stresses 
is  now  given  by  the  rule  above,  viz. :  (1)  and  (12) 
compression,  (2)  and  (11)  tension.  That  is,  the 
members  are  alternately  in  tension  and  compres- 
sion. 

The  reaction  or  shear  in  (1)  is  strictly  14,082J^. 
If  this  correct  value  is  taken,  the  shears  in  (1)  and 
(12),  (2)  and  (11),  or  members  equally  distant 
from  the  centre,  are  all  equal.  Hence,  the  stresses 
in  them  are  equal;  and  we  have  seen,  too,  that 
they  are  of  the  same  character. 

This  must  likewise  follow,  because  the  truss  and 
loads  are  symmetrical  about  the  centre. 

Example  II. — Suppose  sections  passed  through 
diagonals  (4)  and  (9),  construct  the  two  figures  to 


§13]  26 

left  of  section  and  supply  forces  as  in  Arts.  10  and 
11  (figs.  3  and  4),  find  value  and  character  of 
stresses  in  diagonals  (4)  and  (9)  when  sec.  i  =  1.25. 
Bo  the  same  for  diagonals  (5)  and  (8). 

POSITION  OF  LIVE  LOAD,  GIVING  MAXIMUM 
POSITIVE  SHEAR  IN  A  WEB  MEMBER. 

13.  The  case  of  actual  wheel  loads  will 
be  taken  up  later.  In  this  article  the 
uniform  live  load  per  panel  will  be  re- 
garded as  concentrated  at  the  apices  of 
the  chord  which  bears  the  roadway.  Fur- 
ther on  the  actual  position  of  the  moving 
load  will  be  considered. 

The  shear  for  equal  live  apex  loads 
is  a  positive  maximum  when  the  apex 
loads  extend  from  the  farthest  right  abut- 
ment to  the  section  cutting  the  web  mem- 
ber, with  none  of  the  loads  lying  on  the 
left  of  the  section. 

Referring  again  to  fig.  1,  the  largest 
positive  shear  occurs  in  the  diagonal  cut 
by  the  section  d  e,  when  the  three  lower 
apices  to  the  right  of  the  section  are 
loaded,  each  with  a  panel  live  load. 

(a)  For  as  the  shear  in  the  cut  diagonal 


27  [§13 

is  R —  2  w  (Art.  10) ;  if  any  apex  load  on 
the  part  B  d  e  is  taken  off,  R  is  diminished, 
but  2  w  (the  sum  of  the  loads  left  of  the 
section)  remains  the  same ;  hence  the  shear 
is  less  than  when  such  apex  to  the  right  of 
the  section  is  loaded. 

(b)  Again,  if  any  apex  live  load  is  added 
to  the  left  of  the  section  d  e,  R  is  increased 
by  apart  of  it  only,  whereas  2  w  is  aug- 
mented by  the  whole  of  it,  so  that  the 
shear  R  —  2  w  is  less  than  before.  We  con- 
clude as  enunciated. 

If  we  suppose  the  truss  temporarily 
without  weight  with  live  apex  loads  on 
the  left  of  the  section,  thus  covering  the 
shorter  segment  and  none  on  the,  right,  the 
shear  is  negative,  as  seen  in  case  (#),  and 
a  maximum.  As  the  shear  along  d  e>  due 
to  dead  load,  is  plus  (as  d  e  is  to  the  left  of 
the  centre)  the  two  shears  conspire  against 
each  other,  whereas  when  the  longest 
segment  alone  is  loaded,  the  shear  due  to 
both  live  and  dead  loads  is  positive  and 
their  sum  is  the  total  shear  We  shall  see 
later  on  the  significance  of  these  remarks. 


§14]  28 

WARREN  TRUSS  WITH  SUSPENDERS. 

14.  Figure  5  represents  a  triangular 
truss  with  parallel  chords  of  180  ft.  span 
divided  into  12  panels  of  15  feet  each.  The 


E      J      (7      4-      £'    5      C'     b      A 

Fig.  5. 

height  of  truss,  fig.  5,  centre  to  centre  of 
chords,  is  assumed  at  26  feet,  giving  the 
length  of  a  diagonal  30  feet  and  sec.  i  — 
30  -r-  26  ==  1.154. 

If  this  is  regarded  as  a  highway  bridge, 
20  ft.  between  trusses  in  the  clear,  the  dead 
load  per  linear  foot  on  one  truss  is  given 
by  (2)  Art.  5,  as, 

70  -f  6  b  +  0.1  b  I  —  0.2  /, 
or  for  b  =  20, 1  =  180,  514  pounds  per  foot. 
The  panel  load  is  thus,  514  X  15  =  7710 
Ibs.,  of  which  f  or  5140  Ibs.,  goes  to  each 
lower  apex  and  £  or  2570,  to  each  upper 
apex. 

The  live  load  will  be  assumed  at  93J 
pounds  per  square  foot  of  roadway,  which 


29  [  §  15 

gives  |(15  X  20+ 93f):=  14,000  pounds  for 
the  panel  or  apex  live  load,  which  acts 
only  at  the  apices  of  the  lower  chord. 

The  vertical  suspenders  at  B,  D,  F 

simply  hold  up  5140  Ibs.  dead  and  14,000 
Ibs.  live  load  when  fully  loaded. 

The  vertical  posts  at  C.  E ,  are  only 

intended  to  support  the  upper  chord,  and 
hence  sustain  2570  Ibs.,  though  it  is  really 
less  than  that,  as  the  upper  lateral  bracing 
will  be  supposed  connected  to  the  upper 

chord  only  at  points  B,  D,  F ,  and 

hence  the  vertical  posts  do  not  get  the  full 
allowance  of  £  the  dead  panel  load.  It  re- 
mains now  to  compute  the  stresses  in  the 
diagonals  and  chords. 

15.  Diagonals.  The  left  reaction  due 
to  dead  load  acting  at  11  lower  apices  and 
11  upper,  is  7710x11-  2  =  42,405,  and  this 
is  the  dead  load  shear  in  A  B.  By  sub- 
tracting the  sum  of  the  loads  at  apex  B 
and  the  apex  vertically  under  ita  or  7710 
Ibs.,  we  get  the  shear  in  B  C.  Continuing 
thus  to  subtract  7710  from  each  shear  in 
turn,  we  find  all  the  dead  load  shears 
given  in  the  table  below. 


§  15  ]  30 

The  reaction  due  to  the  live  load  when 
it  covers  11  lower  apices,is  14,000  X 11-5-2= 
77,000  Ibs.,  the  live  load  shear  in  A  B.  It 
is  most  convenient  in  what  follows  to  use 
a  panel  length  as  the  unit  of  length. 

For  the  maximum  shear  in  B  C,  by 
Art.  13,  the  live  load  must  extend  from 
the  farthest  abutment  to  C.  The  reaction, 
which  is  the  shear  for  live  load  only,  is 
less  than  before  by  the  amount  of  reac- 
tion due  to  the  apex  load  to  left  of  C, 
which  has  been  taken  off,  or  H  (14,000)  = 
12,833,  giving  the  reaction  64,167  Ibs., 
which  is  the  maximum  shear  due  to  live 
load  only  in  B  C. 

For  C  D  we  similarly  take  off  load  at  C, 
whose  reaction  if  (14,000)  =  11,667  must  be 
subtracted  from  preceding  shear  64,167  to 
get  the  shear  52,500  Ibs.  in  C  D. 

Continuing  thus,  as  the  load  backs  off  to 
the  right,  we  subtract  in  order, 

T9Y,    TV,    TV,    TV,  •  •  •  •  ,  iV 

of  14,000,  and  thus  get  the  live  load  shears 
in  the  table. 


31 


[§15 


1 

rT 

„ 

1    « 

o 

1 

1 

£ 

^  1   j 

^. 

N 

8 

• 

5 

*o    *   o 

i3    +s          ,5 

^    S 

r£      QC 

02   02 

*0                   |                JS    H 

J     ^ 

3 

|  a4  1 

o 

g                                      | 

3 

H 

1  nJl 

O    tt 

1 

AB 

+42,405 

+77,000 

+119,405 

C 

C 

BC 

+34.695 

+64,167 

+  98,862 

t 

2 

CD 

+26,985 

+52,500 

+  79,485 

c 

E 

DE 

+19,275 

+42,000 

+  61,275 

t 

3 

EF 

+11,565 

+32?667 

+  44,232 

c 

G 

F  G 

+  3,855 

+24,500 

+  28,355 

t 

4 

F  G 

-  3,855 

+17,500|+  13,645 

c 

E 

E  F 

-11,565 

+11,667 

+        102 

t 

5 

D'E 

-19,275 

+  7,000 

—  12,275 

t 

C 

CD' 

—26,985 

+  3,500—  23,485 

c 

6 

B  C 

—  34,695 

+  1,167—  33,528 

t 

A 

A  B 

—42,405 

+          o—  42,405 

c 

The  total  shears  are  found  by  adding 
algebraically  those  due  to  dead  and  live 
loads  respectively.  The  character  of  the 
stress  is  given  by  the  rule  in  Art.  12. 
The  live  load  shears  being  formed  here, 
by  subtracting  successively  li  (14,000), 
if  (14,000), from  the  preceding  shear, 


§  16  ]  32 

when  the  result  for  A1  B1  comes  out  zero, 
it  shows  that  the  work  is  correct,  as  no 
live  load  is  then  on  the  bridge. 

These  subtractive  terms  can  easily  be 
found  by  adding  ^  (14,000)  =  1167  to 
itself  to  get  T2T  (14,000),  then  1167  to  this 
to  get  -fa  (14,000),  and  so  on. 

Throwing  away  decimals  will  always 
cause  such  results  to  differ  a  few  pounds 
from  the  true  ones,  which  is  of  no  conse- 
quence in  practice. 

If  desired,  any  live  load  shear  can  be 
found  independently.  Thus,  to  get  the 
shear  in  F'  G  due  to  loads  of  14,000  Ibs. 
each  at  4,  E',  5,  C',  6 ;  note  that  the  re- 
sultant of  the  five  loads  acts  at  5.  Hence 
the  left  reaction  is  ft  of  (5  X  14,000)  = 
17,500,  and  this  is  the  live  load  shear  in 
F'  G  for  this  position  of  the  load. 

The  total  shears  given  in  the  table  for 
the  diagonals  left  of  the  centre  are  maxi- 
mum shears,  as  the  live  load  extends  from 
the  farthest  (right)  abutment  to  the  diag- 
onal considered. 

16.  Counter- Braces. — By  the  rule  of 
Art.  12,  when  the  live  load  extends  from 


33  [§16 

A'  to  4,  since  the  shear  in  F'  G  is  + 
13j645,  this  piece  is  in  compression  and 
subjected  to  a  stress  of  13,645  sec.  i  — 
15,746  pounds. 

Similarly,  when  the  live  load  extends 
from  A7  to  E',  the  diagonal  E'  F'  is  under 
a  tension  of  102  sec.  i  =  118  pounds. 

If  any  of  the  live  load  was  taken  off  to 
the  right  of  the  sections  cutting  these  mem- 
bers, or  any  added  to  the  left  of  the  sec- 
tions, the  live  load  +  shear  is  less  (Art.  13) ; 
and  hence,  since  the  dead  load  shear  is 
minus,  the  total  shear  if  positive  would 
be  less.  The  maximum  compression  that 
F'  G  can  ever  experience  is  thus  13,645 
sec.  i  pounds,  and  the  maximum  tension 
that  E'  F'  can  ever  sustain  is  102  sec.  i 
pounds.  Now,  since  F'  G  and  F  G  are 
symmetrically  situated  with  respect  to  the 
centre  G,  when  the  live  load  extends  from 
A  to  3,  F  G  must  sustain  a  compression  of 
13,645  sec.  i,  the  same  that  F'  G  sustains 
when  the  live  load  extends  from  A'  to  4. 

The  piece  F  G  must  then  be  designed 
to  sustain  not  only  the  maximum  tension 
of  28,355  sec.  i  pounds,  given  in  the  table, 


§17]  34 

when  the  live  load  extends  from  Ar  to  G, 
but  also  a  maximum  compression,  of  13,645 
sec.  i  pounds  when  the  live  load  extends 
from  A  to  3. 

Similarly,  the  diagonal  E  P  must  not 
only  be  designed  to  sustain  the  compres- 
sion given  in  the  table,  of  44,232  sec.  i 
pounds,  when  the  live  load  extends  from 
A'  to  3,  but  likewise  a  tension  of  102  sec.  i 
when  the  live  load  extends  from  A  to  E, 
for  the  last  stress  must  equal  that  in  E'  F', 
in  amount  and  character,  for  a  live  load 
extending  from  A'  to  E',  since  the  mem- 
bers E  F  and  E'  F'  are  symmetrically 
situated  as  to  the  centre  G.  The  diagonals 
E  F,  E'  F',  F  G,  F'  G'  are  thus  "  counter- 
braces"  (Art.  3),  and  we  shall  find  that 
these  are  the  only  members  in  the  truss 
requiring  counter-bracing. 

17.  Minimum  Stress  in  a  Web  Mem- 
ber.— Recurring  to  the  table,  we  see  that 
diagonals  D'  E',  C'  D',  B'  C'  and  A7  B', 
when  the  live  load  extends  from  A'  to  5, 
C',  6  and  A',  respectively,  receive  the  same 
character  of  stress  as  when  subjected  to 
dead  load  only,  which  likewise  gives  a 


35  [§17 

minus  shear  for  each  of  them,  as  given  in 
the  table.  These  positions  of  the  live 
load  (extending  to  the  nearest  abutment) 
cause  the  minimum  stress  the  members  in 
question  ever  receive,  for  the  -f-  shear  is 
greatest  for  the  positions  given,  since  any 
live  load  taken  off  the  right  of  the  section 
or  added  to  the  left  diminishes  the  -[-shear 
(Art.  13),  and  the  greater  the  +  shear  the 
less  the  "total" — shear,  as  we  see  from 
the  table. 

Now,  from  considerations  of  symmetry, 
diagonal  B  C  with  live  load  from  A  to  1 , 
receives  the  same  tension  as  B'  C'  with 
live  load  from  A'  to  6.  Similarly,  C  D 
receives  the  same  compression  with  live 
load  from  A  to  C,  as  C'  D',  when  the  live 
load  extends  from  A'  to  C',  and  in  the 
same  way,  minimum  stress  in  D  E  =^ 
stress  in  D'  E',  as  found  from  the  table. 

We  can  now  make  out  the  table  for 
maximum  and  minimum  stresses  in  the  web 
members,  together  with  the  extremes  of 
stress  in  the  counter-braces,  by  multiply- 
ing the  shears  in  the  table  above  by  sec. 
i  =  1.154. 


§18] 


36 


A  B 

BC 

CD 

D  E 

Max.  .  . 

Min... 

—  137,793 
—    48,9:55 

+  114,087 
-f-     38,691 

—    91,726 
—     27,102 

4-     70.711 
+     14,165 

COUNTER-BRACES. 


EF 

FG 

Max 

51  043 

-f-  32  722 

Min. 

_j_    ny 

i5  746 

The  plus  sign  is  here  used  to  indicate 
tension;  the  minus  sign,  compression. 

18.  If  there  should  be  any  difficulty  in 
comprehending  the  subject  of  counter- 
braces  and  minimum  stress,  it  may  be  en- 
tirely removed  if  the  reader  will  make  a 
rough  sketch  of  the  truss,  large  enough  to 
mark  alongside  each  diagonal  the  amount 
of  the  total  shears  (without  sign)  given  in 
the  first  tablo  above  and  the  character  of 
the  stress,  t  for  tension,  c  for  compression. 

These  shears  correspond  to  the  live  load 
extending  from  the  right  abutment  to  the 
diagonal  in  question. 

Then  suppose  the  live  load  to  extend 
from  the  left  abutment  to  the  diagonal 
considered  and  again  mark  the  shears 


37  [§18 

(without  sign)  on  each  diagonal  and  the 
character  of  the  stress  as  given  in  Arts.  16 
and  17.  Or,  if  preferred,  we  can  suppose 
the  truss  turned  end  for  end,  so  that  A'  is 
at  A  and  A  at  A'.  Then  as  the  live  load 
backs  off  to  the  right,  the  same  total 
shears,  on  the  diagonals  at  the  left  end  of 
the  live  load,  will  be  experienced  as  before, 
for  members  equally  distant  from  the  left 
abutment.  Mark  these  total  shears  (with- 
out sign)  and  the  character  of  the  stress 
from  the  first  tracing  above,  on  a  second 
drawing  placed  below  the  first.  Thus  the 
maximum  shear  in  C'  D'  of  the  second 
drawing  is  now  79,485,  the  same  previously 
marked  on  C  D  the  member  just  above  it 
on  the  first  drawing  and  the  piece  is  in 
compression  as  previously  marked  on  C 
D.  Turn  the  second  drawing  end  for  end 
and  then  write  the  shears  and  character  of 
the  stress,  as  taken  from  it,  on  the  mem- 
bers just  above  on  the  first  drawing. 
These  last  shears  thus  correspond  to  the 
live  load  on  the  original  truss  backing  off 
the  span  to  the  left. 

It  will  now  be  seen  that  pieces  A  B,  B 


§19]  38 

C,  C  D,  D  E,  and  A'  B ,  B'  C',  C'  D',  D  E' 

are  subjected  to  but  one  kind  of  stress, 
whereas  the  remaining  diagonals  are  sub- 
jected in  one  case  to  tension,  and  in  the 
other  to  compression,  and  hence  must  be 
designed  as  counter-braces. ,  The  first  set 
of  shears  are  experienced  as  the  live  load, 
moving  from  right  to  left,  reaches  each 
diagonal  in  turn;  the  second  set  as  the 
live  load,  moving  from  left  to  right,  reaches 
each  diagonal  in  turn.  The  stresses  are 
those  given  in  the  second  table  above. 

This  subject  has  been  carefully  given, 
and  should  be  thoroughly  understood  be- 
fore taking  up  other  trusses  where  a 
knowledge  of  it  is  assumed. 

CHORD    STRESSES    BY    THE     METHOD     OF 

MOMENTS. 

19.  To  prove  that  each  apex  of  the 
chord  that  bears  the  roadway  must  be 
loaded  to  cause  maximum  stresses  in  the 
chords,  consider  the  truss  without  weight, 
and  that  a  single  apex  load  w  rests  any- 
where to  the  right  of  the  centre  of  mo- 
ments E  for  finding  the  stress  in  D  F. 


39  [§19 

Pass  a  section  cutting  D  E,  C  E  and  D  F 
and  supply  forces  as  in  Fig.  4,  Art.  11. 
The  reaction  R  gives  a  right-handed  mo- 
ment R  X  A  E  about  E,  which  must  be 
resisted  by  a  left-handed  moment,  hence 
the  force  applied  to  the  cut  chord  D  F  acts 
to  the  left,  giving  compression  in  D  F  and 
its  moment  is  (the  stress  in  D  F)  X  h,  .  • . 
compression  in  D  F  :=  R  X  AE-^-A.  As 
R  increases  with  every  load  added  to  the 
right  of  E,  so  does  the  compression  in 
DF. 

Next  consider  apex  loads  to  left  of  E, 
from  E  to  A'  being  unloaded.  The  part 
of  the  truss  left  of  the  section  may  now 
be  removed  and  forces  applied  to  cut  parts 
equal  to  the  forces  caused  by  the  left  part 
on  the  cut  members. 

The  moment  of  forces  to  right  of  sec- 
tion about  E  is  simply  R:  X  A'  E,  where 
Rt  is  the  right  reaction.  This  left-handed 
moment  must  be  resisted  by  a  right-handed 
moment  equal  to  force  applied  to  chord  D 
F  at  the  section,  multiplied  by  h.  The  last 
force  must,  therefore,  act  to  right,  which 
causes  a  compression  in  chord  D  F  =  Rt  X 


§  20  ]  40 

A'  E  -r-  h.  As  this  increases  with  Rt  the 
segment  A  E  must  be  fully  loaded  for  a 
maximum  compression  in  D  F.  There- 
fore, we  conclude  that  any  load  on  the 
bridge  (live  or  dead)  causes  a  compression 
in  D  F,  and  that  the  latter  is  a  maximum 
when  the  bridge  is  fully  loaded. 

Similarly,  it  can  be  shown  that  any 
load  on  the  bridge  causes  tension  in  the 
lower  chord,  which  is  a  maximum  for 
the  bridge  fully  loaded. 

20.  To  find  the  chord  stress  in  any  panel 
of  upper  or  lower  chord,  due  to  the  dead 
load  and  a  full  live  load,  pass  a  section 
cutting  only  three  pieces;  the  chord  piece 
in  question,  a  diagonal  and  the  other 
chord,  as  in  figs.  3  or  4  (arts.  9  and  11), 
and  suppose  the  right  part  of  the  truss  re- 
moved. Then  supplying  forces  equivalent 
to  the  forces  the  right  part  exerts  upon  the 
cut  members,  we  have  a  system  of  forces 
in  equilibrium  acting  upon  the  part  of  the 
truss  left  of  the  section. 

The  simple  rule  now  is:  To  find  the 
stress  in  either  chord  piece  cut  by  the  sec- 
tion, take  as  a  centre  of  moments  the  in- 


41  [  §  20 

tersection  of  the  web  member  severed 
and  the  other  chord,  and  then  express  that 
the  moment  of  the  stress  in  the  chord 
equals  the  algebraic  sum  of  the  moments 
of  the  other  forces  acting  on  the  part  of 
the  truss  considered.  The  upper  chord 
will  be  assumed  in  compression,  the  lower 
in  tension,  as  found  above. 

The  forces  acting  on  the  other  chord- 
piece  and  diagonal  pass  through  the  cen- 
tre of  moments,  hence  their  moments  are 
zero;  therefore,  we  have  simply  to  find 
the  moments  due  to  the  reaction  and  loads 
left  of  the  section  and  put  their  algebraic 
sum  equal  to  the  moment  of  the  chord 
stress.  The  live  and  dead  loads  on  the 
same  vertical  line  have  the  same  arm, 
hence  we  shall  call  P  the  sum  of  the  live 
and  dead  panel  loads  at  any  lower  apex 
and  the  apex  vertically  over  it,  .  * .  P  — 
14,000  +7710  =  21,710  pounds.  The 
panel  length  d  —  15  feet  and  the  height  of 
truss,  centre  to  centre  of  chords,  is  h  —  26 
feet.  The  reaction  due  to  11  loads  of 
21,710  Ibs.  each  is  R  =  119,405  pounds. 

Upper  Chord  Stresses. — B  D.  Pass  section 


§20]  42 

cutting  B  D,  either  B  C  or  C  D  and  the 
lower  chord  and  take  C  for  centre  of  mo- 
ments; then,  stress  in  B  D  X  h  —  R  X  2 

d—  p  x  a. 

D  F.  centre  moments  at  E,  stress  in  D 
F  X  ft  =  R  X  4  <7  —  3Px  2  d,  since  the 
resultant  of  P  at  1,  C  and  2,  is  3  P  act- 
ing at  C,  2  c?  to  left  of  E. 

F '  F' '.  Centre  of  moments  at  G,  stress  in 
F  F'  X  ft  =  R  X  6  d  —  5  P  X  3  d, 
since  P  at  1,  C,  2,  E  and  3,  gives  a  result- 
an  5  P  at  2,  on  account  of  symmetry 
of  loads  about  2. 

For  the  lower  chord: 

A  C.  Moments  about  B. 

Stress  in  A  C    X   h  —  R   X   d. 

C  E.  Moments  about  D. 

Stress  inCE    X   ft  =  R  X  3  d  —  P 
(1  +  2)  d. 

E  G.  Moments  about  F. 

Stress  inEGxft=:RX    5  d  — 

P  (1  +  2  +  3  +  4)  d. 
Substitute  R  =  5.5  P  in    all  the  above 
equations  and  simplify. 


43 


On  substituting  numerical  values  we 
find  the  stresses  in  the  chord  members  as 
follows,  the  upper  chord  being  in  com- 
pression, the  lower  in  tension: 

I      STRESS.      i         A,       !       A* 


AC 

68,893 

BD 

125,260 

56,367 

12,526 

CE 

169,101 

43,841 

12,526 

DF 

200,416 

31,315 

12,526 

EG 

219,205 

18,789 

12,526 

FF' 

225,468 

6,236 

On  finding  the  differences,  /\1?  between 
successive  chord  stresses,  and  taking  the 
difference  of  these  successive  differences, 
we  note  that  they  are  constant,  which 
proves  the  numerical  work,  as  will  be 
shown  later  on  (Art.  23). 

CHORD     STRESSES     BY      THE     INCREMENT 
METHOD. 

21.  Another  method  for  finding  chord 
stresses,  applicable  only  to  horizontal 
chords,  will  now  be  given. 


§21]  44 

For  full  live  and  dead  load,  call,  as  be- 
fore, the  sum  of  the  live  and  dead  apex 
loads  at  any  lower  apex  and  the  apex  dead 
load  at  the  upper  apex  vertically  over  it 
—  P  =  21,710  pounds,  and  find  the 
shears  in  the  diagonals  by  subtraction. 
Thus  shear  in, 

A  B  =  5.5  P  D  E  =  2.5  P 

B  C  =  4.5  P  E  F    =  1.5  P 

C  D  =  3.5  P  F  G  =  0.5  P 

Now  pass  a  (curved)  section  cutting 
members  meeting  at  B  (fig.  6),  supply 


Fig.  6 


forces   at   the   cut   sections   equal   to  the 
forces  caused  by  the  balance  of  the  truss 


45  [§21 

on  these  members,  and  let  the  balance  of 
the  truss  be  removed.  We  have  thus  a 
force  acting  up  along  A  B7  whose  vertical 
component  is  5.5  P;  a  vertical  force  acting 
down  along  B  1 ;  a  force  acting  down 
along  B  C,  whose  vertical  component  is 
4.5  P,  and  a  force  acting  to  left  along  B 
D.  The  directions  of  the  forces  follow 
from  the  fact  that  A  B  and  B  D  are  in 
compression,  B  C  and  B  1  in  tension  (Art. 
12). 

Since  we  have  a  system  in  equilibri- 
um, the  sum  of  the  horizontal  compo- 
nents is  zero;  but  the  horizontal  com- 
ponent of  any  diagonal  whose  ver- 
tical component  is  V,  is  V  tan  i, 
whero  i  is  the  angle  it  makes  with  the 
vertical,  .*.  stress  in  B  D  =  (5.5  P 
-f-  4.5  P)  tan  i  =.  10  P  tan  i.  Similarly, 
pass  a  curved  section  cutting  all  members 
assembling  at  D,  supply  forces,  etc.,  as  be- 
fore. The  force  equal  to  the  compression 
in  the  chord  D  F  acts  to  the  left;  that 
acting  on  D  B  acts  to  right  and  is  equal  to 
the  compression  in  B  D  —  10P  tan  i,  just 
found.  The  horizontal  components  of 


§  21  ]  46 

forces  applied  to  diagonals,  both  act  to 
right,  since  C  D  is  in  compression  and  D 
E  in  tension  (Art.  12).  Hence  by  law 
III.  (Art.  7),  stress  in  D  F  = 

(10  +  3.5  -f  2.5)  P  tan  i  =  16  P  tan  i. 

We  proceed  similarly  at  joint  F,  giving 
stress  in  F  F'  = 

(16  +  1.5  +  0.5)  P  tan  i  =  18  P  tan  i. 

The  chord  stress  is  here  a  maximum 
where  the  shear  is  zero. 

A  similar  method  applies  at  apices  A,  C 
and  E,  of  the  lower  chord. 

Note  that  at  A,  the  force  supplied  along 
B  A  acts  down  (lig.  6)  .*.  tension  in  A  C 
=  5.5  P  tan  i. 

At  joint  C  (fig.  6),  the  force  supplied 
along  D  C  acts  down,  that  along  B  C  up, 
and  the  one  (5.5  P  tan  i)  just  found  along 
A  C,  to  the  left;  since  DC  is  in  com- 
pression and  both  B  C  and  A  C  are  in  ten- 
sion. The  horizontal  components  of  all 
these  forces  therefore  act  to  the  left  .'. 
tension  in  C  E  — 

(5.5  +  4.5  -f-3.5)  P  tani  =  13.5  P  tan  i. 


47  [  §  22 

We  have  similarly,  at  apex  E,  tension  in 
EG- 

(13.5  +  2.5  +1.5)  P  tan  i  =  17.5  P  tent. 

We  have  P  tewt  =  21  710  x  fS=  12,525, 
hence  the  stresses  are  as  follows: 

B  D  =  125,250,  A  C  =  68,887, 
D  F  =  200,400,  C  E  ==  169,087, 
F  F  =  225,450,  E  G  =  219,187. 

The  slight  differences  between  these 
values  and  those  found  in  the  preceding 
article  are  due  to  neglecting  decimals  in  the 
provious  case. 

The  chord  stresses  in  members  to  the 
right  of  the  centre,  under  full  load,  are  of 
course  the  same  as  for  members  the  same 
distances  from  the  centre  and  to  its  left. 

22.  The  above  method,  by  chord  incre- 
ments, is  general  for  all  trusses  with  hori- 
zontal chords  and  may  be  stated  as  follows: 

To  find  the  stress  in  any  chord  panel, 
compute  the  shears  in  all  the  web  members 
to  its  left,  including  those  meeting  at  the 
left  end  of  the  chord  panel,  and  multiply 
each  shear  by  the  tangent  of  its  angle  with 


§23] 


48 


the  vertical.  The  sum  of  the  products  is 
the  stress  in  the  chord. 

It  is  usual  to  test  the  stress  thus  found  in 
the  central  panels  by  the  principle  of  mo- 
ments given  above,  and  this  should  never 
be  omitted,  as  it  affords  a  check  on  the 
entire  work. 

23.  Another  method  of  checking  a  part 
of  the  work  can  be  derived  as  follows : 

Write  down  the  coefficients  of  P  tan  i 
for  the  chord  panels,  first  a  lower  panel, 
then  the  next  upper,  and  so  on. 

Find  the  first  and  second  differences, 
and  note  that  the  last  are  all  equal. 


Coeff's  of 
P  tan  i. 

A, 

A* 

AC, 

5.5 

BD, 

10.0 

4.5 

i 

CE, 

13.5 

3.5 

i 

DF, 

16.0 

2.5 

i 

EG, 

17.5 

1.5 

i 

FF, 

18. 

0.5 

49  [  §  24 

The  constant  second  differences  are  thus 
P  tan  i  =.  12,525,  as  found  in  Art.  20. 

24.  The  minimum  stress  in  the  chords 
is  evidently  that  due  to  dead  load  alone, 
and   the   ratio   of  minimum  to  maximum 
stress  is  simply  the  ratio  of  a  panel  of  dead 
load    to  a   panel  of   dead  and   live   load. 
Thus  this  ratio,  in  this  case,  is   7710  -f- 
21,710  =  0.  355. 

This  ratio  is  useful  in  fixing  the  unit 
stress  to  allow  in  the  chords,  as  the  unit 
stress  should  be  made  less  the  greater  the 
extremes  of  stress. 

FORMULAS  FOR  CHORD  STRESSES. 

25.  Figures  7  and  8  represent  two  trusses 
with  horizontal  chords.     The  first  is  called 
a  Pratt  truss  and  the  second  a  Howe  truss. 
The   first   is  supposed   a  truss  of  a  deck 
bridge,  the  latter  of  a  through  bridge ;  but 
we  shall  see  that  the  formulas  deduced  for 
chord  stresses  below  apply  equally  if  the 
first  belongs  to  a  through  bridge  and  the 
last  to  a  deck  bridge. 

Let  the  trusses  be  loaded  with  their  own 
weight,  supposed  concentrated  at  upper 


25] 


50 


and  lower  apices  according  to  a  suitable 
convention,  and  also  with  a  uniform  live 


fa 


...  nd  ~->l (N-n)  d  £ 

I        2       3       4.    a<  ff      6 


23*j*V 


S* nd  ~  -->& *'  ( M-n)  d 

sa !  4     5 


2      3 


load  concentrated  at  the  apices  of  the 
chord  that  bears  the  roadway. 

Call  P  ~  total  dead  and  live  load  con- 
centrated at  any  apex  and  the  one  just 
above  it. 

!N"  =  total  number  of  panels  (12  in  the 
figures). 

n  =  number  of  panels  from  left  abut- 
ment to  centre  of  moments. 

h  =  height  of  truss,  centre  to  centre  of 
chords. 

d  =  panel  length, 


51  [  §  25 

tn  —  stress  on  nth  panel  of  lower  chord, 
cn  =:  stress  in  nth  panel  of  upper  chord. 
The  dotted  diagonals  in  the  four  central 
panels  are  called  counters,  and  are  not  in 
action  under  a  uniform  load;  for  an  inves- 
tigation similar  to  that  in  Art.  12,  Ex.  I. 
will  show  that  all  the  full  line  diagonals 
in  fig.  7  are  in  tension,  and  all  the  full 
line  diagonals  in  fig.  8  in  compression. 
The  full  line  diagonals  then  in  fig.  7 
will  stretch,  hence  the  dotted  diagonals 
(counters)  will  be  shortened  by  the  dis- 
tortion of  the  panel,  and  as  they  are 
always  of  very  small  section,  being  de- 
signed as  ties  only,  they  can  take  no 
appreciable  compression,  and  hence  may 
be  considered  out  of  action.  The  counters 
in  fig.  8  are  simply  struts,  abutting  at 
their  ends  against  angle  blocks,  and  not 
fastened  to  them  in  any  way;  consequently 
when  the  main  diagonals  are  shortened  by 
the  compression,  the  distortion  of  the 
panel  lengthens  the  distance  between  the 
apices  at  the  ends  of  the  counters,  so  that 
the  latter  are  no  longer  in  action.  Leav- 
ing out  the  counters,  therefore,  we  conceive 


§  25 ]  52 

the  truss  cut  in  two,  along  the  dotted  lines, 
through  the  panel  marked  4  of  the  upper 
chord  in  both  figures,  and  the  right  part 
of  the  trusses  removed.  Then  apuly 
forces  equal  and  opposite  to  the  resistance 
of  the  cut  members,  and  take  the  centre  of 
moments,  for  the  forces  acting  on  the  left 
part  at  £,  the  intersection  of  the  diagonal 
and  lower  chord  to  find  the  compression 
cn  in  the  4th  panel  of  the  upper  chord. 

As  the  force  cn  applied  to  the  cut  top 
chord  acts  to  the  left  (the  top  chord 
being  always  in  compression,  Art.  19),  its 
moment  cn  h,  is  left  handed,  and  it  must 
equal  the  right-handed  moment  of  the  left 
reaction  and  loads  to  the  left  of  b. 
Noting  that  the  distance  A  a  is  n  d,  the 
moment  of  the  left  reaction  R  about  by  is 
R  n  d,  n  in  this  case  being  4.  The  sum  of 
the  loads  between  A  and  a  b  is  (n— 1)  P, 
and  it  is  seen  that  their  resultant  acts  at  i 
n  d  from  a  b,  hence  the  moment  of  their 
resultant  about  b  is  (n  —  1)  P  i  n  <# 

p 

Therefore,  since  R  =  (N — 1)  ^~we  have 


53  [  §  25 

=[(N  —  1)  n  —  (n  —  1)  n]  *  P  rf 
Pr7 

.-.  «B  =  (Br-«)«  ££ (10) 

This  formula  was  deduced  for  n  =  4,  but 
it  is  equally  true  for  any  value  of  n\  for 
if  we  wish  to  find  c59  for  example,  the 
centre  of  moments  is  horizontally  nd(=6  d) 
from  the  left  reaction,  so  that  its  moment  is 
n  d  R.  There  are  again  (n  —  1)  loads  P  to 
left  of  centre  of  moments,  and  they  are 
symmetrically  disposed  about  a  point  mid- 
way between  A  and  the  centre  of  moments^ 
so  that  the  arm  of  the  resultant  is  k  ndsmd. 
the  sum  of  their  moments  (n  —  1)  P  |  n  d, 
as  before. 

It  is  always  understood,  in  writing  down 
general  formulas  of  this  kind,  that  the 
reader  will  test  their  generality  by  imag- 
ining n  to  increase  or  decrease  by  1,  and 
seeing  that  the  results  are  true,  no  matter 
what  n  may  be. 

To  find  the  stress  £n,  in  the  nih  panel 
of  the  lower  chord,  we  suppose  a  section 
taken  across  the  nih  panel  of  lower  chord, 
say  the  one  marked  4  in  the  figures.  The 


§  25  ]  54 

centre   of   moments  is  now  at  a,  and  the 

forces   and   their   arms  are   given  by  the 

same  formulas  as  above;  therefore  we 
deduce  as  before, 

<n  =  (N  -  n)  n  ?£  ......  (11) 

Therefore,  to  find  the  chord  stresses  in 
panels   marked    1,  2,   ......  ,   of   the   upper 

chord,  or  of  those  marked  on  the  figures3 
1.    2,     ......  ,    of    the     lower     chord,     we 

substitute  successively  ^  =  1,  n  =z  2,  etc.,  in 
formula  (10)  or  (11).  The  upper  and 
lower  chord  panels  similarly  numbered  on 
the  two  figures  thus  have  the  same  stresses, 
a  result  which  follows  from  Art.  9,  eq.  9, 
likewise,  if  we  pass  a  section  cutting  any 
two  chord  pieces,  marked  with  the  same 
number,  and  the  vertical  member  between 
them  and  place  the  sum  of  the  horizontal 
components  equal  to  zero 


Example.  —  As  in  Art.  20,  take  N  —  12,  7i  = 
26,  d  =  15,  P  —  21,710  pounds.  On  substituting 
numerical  values  in  (10)  and  (11),  we  find  the 
same  chord  stresses  given  in  Article  20.  In  fact, 
the  formulas  can  easily  be  proved  to  apply  to  Fig. 


55  [  §  2(> 

5,  Art.  20,  if  we  number  the  upper  chord  panels 
B  D,  D  F,  F  F',  2,  4,  6  and  the  lower  AC,  C  Ey 
E  G,  1,  3  and  5,  and  thus  find  for  the  upper  chord 
cz,  C4»  CQ  from  (10)  and  for  the  lower  chord  t19 
tsi  t5  from  (11).  This  follows,  as  will  be  seen  in 
going  over  the  proof  of  the  last  article,  because 
n  d  will  always  give  the  arm  of  the  reaction  for 
the  chord  panel  marked  n,  and  the  (n  —  1)  loads 
P  are  similarly  situated  in  all  three  trusses. 

Prove  the  formulas  (10)  and  (11)  applicable  to 
fig.  5  (Art.  20)  in  the  manner  stated. 

26.  From  the  method  of  derivation  of 
eqs.  (10)  and  (11),  it  is  seen  that  they  al- 
ways give  the  stress  in  a  chord  member 
whose  centre  of  moments  is  n  d  from  the 
left  abutment,  the  loads  being  placed  as  in 
figures  7  and  8. 

The  formulas  (10)  and  (11)  can  be  put 
under  another  form  if  we  let  m  —  N  —  n 
=•  number  of  panels  to  right  abutment 
from  the  centre  of  moments,  and  remem- 
ber that  n  —  number  of  panels  from  left 
abutment  to  centre  of  moments,  for  the 
chord-piece  considered. 

4  =  cn  =  n  m  —  (12) 

Example  J.  —  In  the   Warren  girder,  Fig.  1  of 


§27]  56 

Art.  2,  of  6  panels,  let  d  =  h  =  10  and  let  P  -  2 
tons,  represent  the  panel  live  load  at  each  apex  of 
the  lower  chord,  the  dead  load  not  being  consid- 
ered. 

We  can  find  by  (12)  above  the  stresses  in  the  top 
chord,  but  not  in  the  lower.  Why  ? 

For  first  upper  chord  panel,  the  centre  of  mo- 
ments is  one  panel  length  from  A  or  n  =  1,  .  *  . 
cl  =1X5  =  5  tons. 

For  second  panel,  centre  is  2  d  from  A  or  n  —  2, 
,  ' .  C2  =  2  X  4  =  8  tons. 

For  third  panel,  cs  =3X3  =  9  tons. 

Find  by  the  method  of  moments  of  Art.  20 
these  values,  also  the  lower  chord  stresses,  tL  = 
2.5,  £g  =  6.5,  t3  =  8.5. 

Example  II. — Find  by  the  method  of  chord  in- 
crements (Art.  22)  the  same  chord  stresses  in  last 
erample. 

27.  SECOND  DIFFERENCE  CONSTANT. 

After   computing   the  chord  stresses  in 

panels  1,  2,  3, .  .  .  ,  in  turn  (fig.  7  and  8),  it 

is  well  to  check  by  taking  first  and  second 

differences,  as    in    Art.    20.     The  second 

differences  are  constant  and  equal  to  P  — 
Thus: 

/XT  X         Pd 

cn  =  (N-n)n  — 


57  [§28 

cn  +  1  =  (N-n  — 1)  (n-f-1)  J| 

Pe? 

cn  +  1  —  cn  =  =  (N  —  2  n  —  1)    — 

This  first  difference  changes  (2)  —=-  —  P  — 

for  each  increase  of  ?£  by  1,   /.  the  second 
difference  is  constant. 

In   the    example   given  in  Art.  25  the 

second  difference  P  7-  —  12,525  as  before 
h 

found. 

WARREN  TRUSS. 

28.  The  chord  stresses  in  the  Warren 
Deck    Truss,    fig.     9,    having    diagonals 

If    $    i>    v    ^ 

H     B  D  F  D1  B'    H' 


A 


C  £ 

ffk~~ md 


Fig.  9. 

equally  inclined  to  the  vertical,  are  easily 
made  out  by  the  method  of  chord  incre- 


§281  58 

mentf,  Art.  22,  but  we  shall  deduce  a  for- 
mula for  the  tension  in  the  lower  chord 
panels  for  a  uniform  live  load  covering  the 
upper  chord. 

Here  a  full  panel  load,  p,  is  held  up  at 
the  intermediate  apices  D,  F  and  D1  and 
%  p  at  the  end  apices  B  and  Bl  (why  ?)  It 
is  convenient  in  taking  moments,  to  regard 
the  load  at  B  as  replaced  by  p  acting  down 
and  J-  p  acting  up. 

To  find  the  chord  stress  =  t,  in  E1  C1, 
take  D1  as  a  centre  of  moments.  Call  m 
the  number  of  panel  lengths  and  fractions 
from  A  to  1  and  ml  the  number  from  1  to 
A1.  Thus,  m  ~  3£,  ml  =  14,  for  centre  of 
moments  at  D1.  The  distance  A  1  =  m  d 
and  A1  1  =  m1  dy  where  d  is  a  panel 
length.  The  height  of  truss  is  h. 

We  have  for  the  left  reaction, 

R  —  (  m  +  m1  —  *)-£- 

£ 

and  its  arm  about  D1  is  m  d.  The  result- 
ant of  the  forces  acting  down  to  left  of 
D1  =  (m  —  %)p,  and  its  arm  is  4  (m  -}-  J)  d. 


59  [§29 

The  force  i  p  acting  up  at  B  has  an  arm 
(ra  —  i)  d. 

The  sum  of  the  moments  of  these  forces 
about  D1  equals  t  h. 

dp 

.-.  t  h  —  R  m  d  -f-  \p  (in  —  £)  d  —  (m2  —  J)  -^ 


(13) 


29.  The  stress  in  any  upper  chord  panel 
is  the  mean  of  the  stresses  in  the  two  ad- 
jacent lower  chord-panels.  To  prove  this, 
let  Q  =  resultant  of  loads  to  left  of  E1 
and  x  its  distance  from  F;  then  the  mo- 
ment of  all  forces  to  left  of  E1,  about  F, 
divided  by  h  gives, 

The  stress  in  EE'  =  (R.  H  F  —  Q  x)  -^  h 
Similarly  the  stress  in  E'  C'  is 

[R.  H  D'  --  Q  (x  +  d)]  -  h 
The  mean  of  these  two  stresses  is, 

[R.  A  E'  —  Q  (x  +  i  d)}  -*-  h, 
which  is   seen  to    be  the  moment  of   all 
forces  to  the  left  of  E',  about  Er,  divided 
by  7i,  and  hence  this  mean  is  the  stress  in 
F  D  .  Q.  E.  D, 


§30]  60 

30.  Example  I. — In  the  5-panel  deck  truss  of 
fig.  9,  let  d  —  16,  7i  =  10  feet  and  the  live  load  on 
one  truss,  2000  pounds  per  linear  foot,  corre- 
sponding to  a  very  heavy  train  of  railroad  cars. 
This  gives  p  =  32,000  Ibs.,  and 

p  d 
^=25,600. 

Using  formula  (13)  we  find  the  stresses  in  A  C, 
C  E,  and  E  E',  respectively  to  be  57,600,  134,400 
and  160,000  pounds.  The  mean  of  the  first  two, 
96,000  Ibs.,  is,  by  Art.  29,  the  stress  in  B  D,  and 
the  mean  of  the  last  two,  147,200  Ibs.,  is  the  stress 
in  DP. 

Test  these  results  by  the  method  of  chord  in- 
crements. 

Example  II.  — For  a  railroad  bridge,  the  weight 
per  linear  foot  for  the  above  truss  of  80  ft.  span, 
by  eq.  4,  Art.  5,  is  580,  giving  a  weight  per  panel 
of  9280  pounds.  This  weight  should  be  used  in 
practice,  but,  for  convenience  of  computation,  let 
us  take  a  panel  weight  of  9000  Ibs.  and  express  it 
in  thousand  pounds.  That  is,  the  weight  per  panel 
is  9  thousand  pounds,  of  which  one-third,  or  3, 
will  be  the  apex  load  on  the  lower  chord  and  6  on 
the  upper  chord,  except  at  the  ends,  B  and  B1, 
where  the  apex  load  will  be  assumed  f  6  =  4. 5. 

Draw  the  figure  of  the  truss  fig,  9  and  mark  OIL 
the  diagonals  the  resulting  shears:  A  B  =  19.5, 
B  C  =  15,  C  D  =  12,  D  E  =  6,  and  E  F  =  3 ; 


61  [§30 

then  by  the  method  of  chord  increments,  Art.  22, 
find  the  chord  stresses : 

A  C    =  15.6,  BD  =  27.6 

CE    =  37.2,  DF  =  42.0 

E  E'  =  44.4, 

Test  these  stresses  by  the  method  of  moments. 
Thus  for  C  E,  centre  of  moments  is  at  D  and  B  = 
19.5;  therefore,  stress  in  C  E  equals 

19.5  X  24  —  4.5  X  16  —  3  X  8 

~To~ 

Example  III.— Find  the  maximum  and  mm i- 
rnum  stresses  in  the  diagonals  of  the  same  truss, 
due  to  the  dead  lead  above  and  a  live  load  of  32 
thousand  pounds  per  panel  moving  over  the  top 
chord.  The  shears  due  to  dead  load  are  found  as 
in  the  last  example  and  entered  in  the  following 
table : 

For  live  load  shears  proceed  as  in  Art.  15,  noting 
that  the  apex  loads  at  B  and  B  are  each  J  (32)  = 
24.  The  left  reaction  R  for  full  load  is  thus  72, 
and  this  is  the  maximum  shear  in  A  B. 

The  max.  shear  for  B  C  and  C  D  equals  72,  minus 

4  5 
the  reaction  (24  X~ 5-  -  21.6)  due  to  load  at  B 

removed,  and  so  on  (see  Art.  15).    , 


§30] 


Live      1 
Load 
from 

Members 

Dead  Load 
Shear 

Live  Load 
Shear 

Total 
Shear. 

H  toB 

A  B 

+  19.5 

+  72.0 

+  91.5 

c 

H'toD 

BC 

+  15.0 

+  50.4 

+  65.4 

t 

H'  toD 

CD 

+  12.0 

+  50.4 

+  62.4 

c 

H  toF 

DE 

+    6.0 

+  28.0 

+  34.0 

t 

H'toF 

EF 

+    3.0 

+  28.0 

+  31.0 

c 

H'  to  D 

E'F 

-    3.0 

+  12.0 

+    9.0 

t 

H'  to  D 

D  E 

-    6.0 

+  12.0 

+    6.0 

c 

H'  to  B 

CD' 

—  12.0 

+    2.4 

—    9.6 

c 

H'  to  B 

B  C 

-  15.0 

+    2.4 

—  12.6 

t 

H  to  H 

A  B 

-19.5 

0.0 

-19.5 

c 

The  length  of  a  diagonal  is  >/88  +  108  =  12.8^ 
hence* we  multiply  each  of  the  shears  above  by 
sec.  i  —  12.8  -r-  10  =  1.28  to  get  the  max.  and 
min.  stresses  below  (see  Arts.  16  and  17),  which  are 
given  in  pounds. 


AB 

BC 

CD 

DE 

EF 

Max. 
Min. 

—  117,120 
-    24,960 

+  83,712 
+  16,128 

—  79,872 

—  12,288 

+  43,520 
-    7,680 

—  39,680 
+  11,520 

In  this  table,  the  plus  sign  indicates  tension ;  the 
minus  sign,  compression.  It  will  be  noticed  that 
only  D  E  and  E  F  (and  of  course  D'  E',  E'  F  )  need 
counter-bracing. 


63  [  §  31 

31.   Through  Warren  Girder. 

Let  us  next  consider  a  Warren  truss 
of  five  panels,  lettered  as  in  fig.  9,  but 
with  the  roadway  on  the  bottom  chord. 
The  members  A  H,  H  B,  A'  H',  and  H'  B' 
are  now  omitted.  No  span  or  height  is 
given  in  this  example. 

If  the  total  dead  load  on  one  truss  is  W, 
then  the  dead  load  per  panel  W  -5-  5  will 
be  called  w.  The  live  load  per  panel  will 
be  called  p.  Let  us  suppose  that  the  dead 
panel  load  is  divided  between  the  two 
chords,  so  that  %  w  rests  at  each  lower 
apex  and  14  w  at  each  upper  chord  apex, 
except  at  B  and  B'  where  £  w  is  supposed 
to  rest. 

We  have  now,  for  dead  load.  R  =  2  w 
from  which,  by  subtraction,  we  get  the 
dead  load  shears  on  the  diagonals,  which 
can  be  marked  on  them.  Then  compute 
the  live  load  shears*  and  enter  both  in  the 
following  table.  Call  s  =  secant  of  the 
angle  and  t  =  tangent  of  the  angle,  any 
diagonal  makes  with  the  vertical.  Any 

*  The  live  load  shears  are  most  conveniently  found  by  aid 
of  formula  (14)  of  the  next  article. 


31] 


64 


shear  on  a  diagonal  multiplied  by  s  gives 
its  stress. 

To  find  the  chord  stresses,  apply  the 
method  of  chord  increments,  Art.  22,  and 
place  the  results  for  dead  load  and  also  for 
a  full  live  load  in  the  following  table: 


Live  Load 
from 

Member. 

Stress 

Live  Load. 

Dead  Load. 

A'  toC 

AB 

—  2.  0  p  s 

•—  2.000  ws 

A'  toC 

BC 

+  2.0  p  s 

+  1.875  ws 

A'  toE 

CD 

—  1.2  .ps 

—  1.125  ws 

A  toC 

CD 

+  0.2  ps 

—  1.125  w  s 

A'  toE 

DE 

+  1-2  ps 

+  0.875  w  8 

A  toC 

DE 

—  0.2  ps 

+  0.875  w& 

A'  to  E' 

EF 

-  0.6  p  s 

—  0.125  w  s 

A  toE 

EF 

+  0.6  p  s 

-  0.125  ws 

A'  to  A 

AC 

+  2.0  pt 

+  2.000  wt 

« 

CE 

+  5.0p£ 

+  5.000  wt 

« 

EE' 

+  6.0  pt 

+  6.000  wt 

tt 

BD 

-4.0  pt 

—  3.875  w  t 

n 

DF 

—  6.0  pt 

—  5.875  wt 

-f  Indicates  tension.    —  Compression. 

The  shear  on  C  D,  when  live  load  ex- 
tends from  A  to  C,  is  the  same  as  the  shear 
on  C'  D'  when  the  load  extends  from  A' 
to  C'.  Similarly  the  second  of  the  live 


65  [ §  32 

load  stresses  for  D  E  and  E  F  were  com- 
puted for  D'  E'  and  E'  F'  for  load  extend- 
ing from  A'  to  C'  and  A'  to  E  ,  respect- 
ively. 

The  maximum  stresses  on  A  B,  B  C  and 
the  chord  members  are  found  by  adding 
the  stresses  due  to  live  and  dead  loads; 
the  minimum  are  due  to  dead  load  alone* 

For  the  other  members  we  find  the 
maximum  stresses  by  adding  the  upper 
expressions  in  the  table;  the  minimum  by 
combining  the  lower,  which  pertain  to  any 
member. 

When  the  span  and  height  are  given  to- 
gether with  w  and  p9  the  stresses  can  all 
be  obtained. 

The  results  in  the  above  table  refer  to 
any  5  panel  Warren  through  bridge,  and 
the  table  is  a  type  of  a  kind  prepared  in 
bridge  offices,  for  instant  use,  for  all  kinds 
of  trusses  and  any  usual  number  of 
panels. 

SHEAR  FORMULAS. 

32.  A  formula  for  shears  for  trusses  of 
the  type  of  fig.  10,  for  a  uniform  live 


§  32  ]  66 

load   over  the  bottom  chord  is  easily  de- 
rived. 

Call  p  —  panel  live  load  for  one  truss, 
d  =  length  of  panel  in  feet, 
U  =  total  number  of  panels, 
n  =  number  of  panels  from  A  to 
the  front  apex  load  (n  =   5    in 
figure). 

By  the  method  of  "  apex  loads,"  which 
will  first  be  given,  the  uniform  load,  half  a 
panel  length  either  side  of  an  apex,  will  be 
regarded  as  concentrated  at  that  apex. 

This  is  true  for  an  apex  in  the  middle  of 
the  load,  but  it  is  impossible  at  the  left 
end.  Thus  at  apex  b,  the  whole  of  the 
half  panel  of  load  to  left  of  b  is  supposed 
borne  at#,  whereas  part  of  it  is  held  up  at 
a  and  the  true  shear  in  panel  5  is  equal  to 
the  reaction  of  the  load,  as  it  rests  on  the 
bridge,  less  this  part  borne  at  a.  In  order 
that  a  full  panel  load  be  borne  to  &,  the 
live  load  must  extend  to  a;  in  which  case 
the  true  shear  is  equal  to  the  left  reaction, 
minus  the  half  panel  load  borne  by  the 
stringers  to  a. 

The  customary   method    of  apex  loads 


67  [§32 

thus  always  gives  an  excess,  except  at  the 
end  panel  (1)  where,  for  full  load,  the 
half  panel  of  load  next  a  (now  coinciding 
with  A)  is  borne  directly  by  the  abutment 
and  the  shear  is  seen  to  be  the  same  by 
either  method. 


Fig.  10. 

By  the  method  of  apex  loads,  there  -are 
(N —  n)  loads  p  at  the  lower  apices,  and  this 
is  true  no  matter  where  6,  is  placed.  Their 
resultant  acts  at  i  B  a  =  i  (N —  n  +  1)  d, 
from  the  right  abutment  B,  wherever  b  is 
placed;  hence  taking  moments  about  B, 
we  have, 

R  N  d  =  (N  —  n)  p%  (N  —  n  +  1)  d 
where  the  left  reaction  =  R. 

Therefore,  as  the  shear  S  in  the  ?*th 
panel  is  equal  to  R,  we  have, 

S  =  (N  —  n)  (N  —  n  +  1)  -. (14) 


§  33  ]  68 

If  we  pass  a  section  cutting  either  of  the 
diagonals  over  the  nih  panel  and  the. 
two  chords,  the  shear  S  by  Arts.  9  and  11, 
is  numerically  equal  to  V,  the  vertical 
component  of  the  stress  in  either  diagonal. 

33.  EXACT  FORMULA  FOR  SHEARS. 

Let  the  live  load  in  fig.  10  extend  a 
distance  x  to  the  left  of  b,  x  being  less  than 
d.  For  convenience,  call  the  distance 
B  b  =  (N  —  n)  d  =  c,  and  consider  a  thin 
slice  of  the  load  at  the  left  end  whose 
weight  =  1  pound.  The  left  reaction  due 

(£         I         ^\ 
—  j-^-  1,  /  being  the  span,  and 
^      / 
the  amount  of  this  1   Ib.  held  up  at  a  is, 

Qft 

-T  ;  hence  the  shear  in  panel  a  b  due  to  it  is, 

c  +  x        x         c  _  (/  —  d)  x 
~l~    "  d  ^    I  dl 

As  x  increases  from  zero,  this  remains 
plus  until  it  attai  ns  the  value  zero,  for  which 


and  for  x  greater  than  this  value  the  shear 
is  minus.     Hence  no  part  of  the  live  load 


69  [§33 

must  extend  to  the  left  of  b  a  greater  dis- 
tance than  the  value  just  found  for  x,  or 
the  shear  will  be  diminished. 

The  distance  x  above,  for  maximum 
shear,  can  be  easily  laid  off  with  dividers 
on  a  drawing  of  the  truss,  as  follows: 
Divide  a  panel  length  into  (N  —  1)  parts 

and  call  the  length  of  one  part  /  ^  -  -  )  n= 

a;  then  for  c  =  d,  2  d,  3  c7,  .  .  .  ,  (N  --  1) 
d,  x  =  a,  2  a,  3  a  .  .  .  ,  dy  which  can  be 
easily  laid  off.  It  is  seen  that  x  increases 
from  x  =.  a  for  c  =  d,  to  x  =  d  or  a  full 
load  covering  the  bridge,  for  the  end 
panel  at  the  left. 

Supposing  that  x  has  the  value  found 
above  for  maximum  shear,  we  have  the 
exact  shear  SA  in  panel  'n  or  b  a,  for  the 
uniform  load  w  per  linear  foot,  extending 
a  distance  c  -J-  x  from  the  right  abut- 
ment, equal  to  the  left  reaction  min  us  the 
load  borne  at  a,  or, 

w  x2  _      w     r  Na  c2  N 

~  ~~    =    ~  ~ 


-  I)2 

e2  ID  dr 


§34] 


70 


In  this  formula  w  d  —  p  =  live  panel 
load  and  (N  —  n)  is  the  number  of  panel 
lengths  from  b  to  B.  Calling  (N  —  n)  — 
m^  (15)  can  be  written, 


m2     p 

Sl  =  N  — 1  2" 


.(16) 


34.  Let  us  compare  the  results  by  (14) 
and  (15)  for  the  five  panel  Warren  truss  of 
Art.  31: 


S  by  (14) 

S  by  (15) 

n  =  l 

2.0  p 

2,000  p 

n  =  2 

1.2  p 

1.125  p 

n  =  3 

0.6  p 

0.500  p 

n  =  4 

0.2  p 

0.125  /> 

The  difference  in  the  shears  by  the  two 
methods  is  seen  to  be  greatest  at  the  cen- 
tre panel,  where  it  amounts  to  0.1  p.  This 
difference  increases  slightly  with  the  num- 
ber of  panels,  so  that  for  a  12  panel  truss 
it  attains  the  value  0.114  p.  It  is,  there- 
fore, small  for  any  usual  value  of  N.  The 
approximate  method  is  thus  seen  to  give  a 
slight  excess  of  shear  in  the  interior  panels 
which  attains  its  maximum  at  the  centre, 


71  [§35 

where  the  section  of  the  web  members  is 
small  and  a  slight  excess  is  not  objection- 
able. 

35.  Formula  for  the  exact  shear  for  a 
Deck  Warren  truss. 

If  in  Fig.  9,  Art.  28,  we  suppose  the 
uniform  load  to  per  foot  to  rest  on  the  top 
chord  a  distance  c  -\-  x  from  H'  where 
c  =.  m  d,  is  the  distance  from  H'  to  an 
apex  arid  m  the  number  of  panels  and  frac- 
tions in  this  distance,  the  reasoning  of  Art. 

y? 

33  leads  to  the  same  value  of  x  =   _=  — 

N  —  1 

and  the  same  formula  for  Sa  in  terms  of  c, 
but  not  in  terms  of  n. 


.  c 


p 


36.  Example.  —  Show,  by  a  procedure  similar  to 
that  of  Art.  27,  that  the  second  differences  derived 
from  the  terms  given  by  formulas  (14),  (15)  and 
(17)  are  all  constant,  and  hence  can  be  used  as  a 
check  on  computation. 

37.  The  formulas  (14),  (15)  and  (16),  for 
a  through  bridge,  give  the  shear  in  the 
web  members  over  the  panel  a  6,  fig.  10, 


§37]  72" 

whether  their  inclinations  to  the  vertical 
are  the  same,  as  in  a  Warren  truss,  or 
different  as  in  Howe,  Pratt  and  other 
trusses;  for  the  inclination  of  the  web 
members  did  not  enter  into  the  derivation 
of  the  formulas.  Hence  we  may  conceive 
the  two  web  members  over  the  panel  a  b> 
fig.  10,  to  have  their  tops  moved  forwards 
or  backwards,  so  that  one  of  them  becomes 
vertical,  giving  trusses  of  the  Pratt  or 
Howe  types  (Art.  25),  and  the  shears  for 
live  load  in  these  are  equal  and  are  given 
by  eq.  (14)  by  the  method  of  apex  loads, 
and  by  eqs.  (15)  or  (16)  by  the  exact 
method. 

Formulas  (14)  and  (17)  apply  to  Pratt 
and  Howe  trusses  when  the  live  load 
passes  over  the  top  chord.  For  such  deck 
trusses,  to  find  the  position  of  the  live  load 
giving  maximum  shear  in  any  tie  or  strut, 
the  simple  rule  is,  pass  a  section  cutting 
only  the  two  chords  and  the  web  member 
(never  more  than  three  members  in  action 
in  all),  and  move  the  live  load  from  the  right 
abutment  up  to  the  apex  just  right  of  the 
section,  if  the  method  of  apex  loads  is  to 


73  [  §  38 

be  used,  or  a  distance  x  =  c  -4-  (N  —  1) 
beyond  this  point  if  the  exact  method  is  to 
be  used.  No  apex  load  should  ever  be 
placed  to  the  left  of  the  section,  or  the 
shear  is  diminished. 

This  rule,  for  deck  trusses  and  for  apex 
loads,  requires  that  for  maximum  shear  in 
a  vertical  post  of  a  Pratt  truss,  the  front 
apex  load  must  be  directly  over  it,  whereas 
for  a  Howe  truss  the  vertical  tie  is  most 
strained  when  the  front  apex  load  is  one 
panel  length  to  its  right. 

The  Pratt  and  Howe  Bridges,  both 
through  and  deck,  will  now  be  taken  up 
in  detail. 

THROUGH  PRATT  TRUSS. 

38.  Let  figure  11  represent  a  through 
Pratt  truss  of  120  feet  span,  divided  into 
6  panels  of  20  feet  each.  The  height, 
centre  to  centre  of  chords,  is  assumed  at 
20  feet.  The  dead  load  at  each  lower 
apex  is  assumed,  for  convenience  in  com- 
putation, at  6,  at  each  upper  apex  3  and 
the  live  panel  load  at  18,  all  loads  being 
expressed  in  thousand  pounds.  The  dead 


§38]  74 

load  reaction  is  22.5  and  we  get  the  shears 
due  to  dead  load,  in  the  diagonals  over 
panels  1,  2,  3  ,  by  the  continued  sub- 
traction of  9. 

Selecting  the  method  of  "  apex  loads," 
formula   (14)  gives  shear  equal  to 

(6  —  n)  (7  —  n)  X  1.5 

From  this,  the  live  load  shears  are  com- 
puted below  for  n  =  1,  2,  ,  and  added 

to  the  dead  load  shears. 


mt  of 
ad  at 

„     Dead  load        Live  load           rp  ,  . 
shear               shear 

c, 

1,     22.5 

+     45 

=     67.5 

E, 

2,     13.5 

+     30 

=     43.5 

G, 

3,       4.5 

+     18 

=?     22.5 

E', 

4,—  4.5 

4-      ^ 

=       4.5 

C' 

5,—  13.5 

+       3 

—  -  10.5 

Find  the  first  and  second  differences  of 
the  total  shears,  and  notice  that  the  second 
differences  are  each  equal  to  3,  a  constant, 
which  verifies  the  work. 

The  dotted  lines  in  the  two  central 
panels  represent  counter  ties  arid  by  Art. 
25,  are  not  in  action  when  the  main  diag- 
onals (ties)  are  in  action.  We  shall  now 


75 


[§38 


take  up,  in  some  detail,  the  maximum  and 
minimum  stresses  ever  experienced  by  the 
web  members. 

Batter  brace  A  B.  The  max.  shear  is 
for  full  load  and  is  67.5 ;  the  min.  shear  is 
due  to  dead  load  only  and  is  22.5 

Hip  vertical  B  C.  Its  max.  stress  is 
simply  one  panel  of  live  load  and  the  dead 
load  held  up  at  C  or  6  +  18  =  24;  min. 
stress  =  6. 

Main  tie  B  E.  The  load  extends  from 
A'  to  E.  The  max.  shear  is  43.5  and  the 
piece  is  in  tension,  as  the  shear  is  plus 
(see  fig.  4,  Art.  11). 

B  D  F          Dr          Bf 


C  E  6     '      £'          C          A' 

Fig. II. 

Main  Tie  D  G.— For  max,  shear,  the 
load  extends  from  A'  to  G,  giving  a  shear 
-j-22.5;  hence  the  tie  D  G  is  in  action 
(Art.  11)  and  the  counter  E  F  is  out  of 
action  (Art.  25). 


§38]  76 

Post  D  E. — The  post  D  E  is  subjected 
to  a  stress  equal  to  the  vertical  component 
of  the  stress  in  D  G  plus  the  dead  load 
held  at  D.  Therefore,  it  receives  its  maxi- 
mum stress  when  the  shear  in  D  G  is  a 
maximum,  or  for  live  load  from  A'  to  G  /. 
max.  stress  in  post  D  E  =  22.5  -f  3  =  25.5. 

Counter  F  E'. — With  live  load  from  A' 
to  E1,  the  shear  is  -f-  4.5,  and  therefore  acts 
up;  hence,  if  we  pass  a  section,  etc.,  as  in 
Art.  11,  V  acts  down;  hence  the  counter  is 
in  action  and  the  main  tie  G  D'  out  of  ac- 
tion, as  it  cannot  take  compression.  The 
max.  shear  in  the  counter  is  thus  4.5. 

Post  F  G. — The  max.  stress  in  F  G  for  load 
from  A'  to  E'  is  the  vertical  component  of 
the  stress  in  the  counter  (4.5)  plus  the  dead 
load  (3)  held  at  F  or  in  all,  7.5.  Its  mini- 
mum stress  is  simply  that  due  to  the  dead 
load  resting  at  the  top  =  3,  when  the  coun- 
ters meeting  there  are  both  out  of  action, 
which  is  the  case  when  there  is  no  live  load 
or  a  full  live  load  on  the  bridge  (why?) 

It  may  be  remarked  that,  when  the  live 
load  extends  from  A'  to  E',  giving  the 
shear  in  panel  4  =  +  4.5,  the  shear  in 


77  §  38  ] 

panel  3  is  greater  by  the  dead  load  held 
up  at  F  and  G;  therefore  the  shear  in 
panel  3  —  +  4.5  -f-  9  =  +  13.5,  and  since 
it  is  plus,  D  G  is  in  action  and  E  F  is 
not  in  action:  so  that  the  counter  E  F 
does  not  transmit  any  stress  to  post  F  G 
for  this  position  of  the  load.  Further, 
this  position  of  the  load  (from  A'  to  E') 
gives  the  maximum  stress  in  both  E'  F  and 
F  G,  for  if  any  load  is  taken  off  or  any 
added  to  left  of  E',  the  positive  shear  is 
diminished  (Art.  13);  in  fact,  if  an  apex 
load  is  added  at  G,  the  shear  in  panel  4  = 
shear  in  panel  3  minus  the  loads  (9  +  18) 
liekl  at  F  and  G  ==  22.5  —  27.  =  -  4.5 
and  the  counter  E'  F  is  thus  out  of  action 
(the  shear  being  minus),  and  F  G  sustains 
only  the  dead  load  at  its  top. 

Post  D'  E  ,  with  the  live  load  extending 
from  A  to  E',  we  have  just  seen  that 
counter  E'  F  is  in  action;  hence  G  D'  is  out 
of  action,  and  cannot  transmit  any  stress  to 
post  D'  E',  which  thus  sustains  only  the 
dead  load  3,  at  its  top.  This  is  the  mini- 
mum stress  in  D'  E',  and  consequently  the 
minimum  stress  in  post  D  E  for  a  live  load 


§  38  ]  78 

from  A  to  E  is  the  same  or  3  thousand 
pounds. 

Tie  B'  E'. — With  one  apex  live  load  at 
C7  the  shear  in  E'  B'  =  —  13.5  from  dead 
load  +  3  from  live  load  =  —  10.5  as  given 
above.  If  any  live  load  is  added  or  taken 
off,  the  •+  shear  is  less  (Art.  13)  which 
would  increase  the  minus  shear  ( —  10.5)  ; 
hence  this  is  the  smallest  negative  shear 
that  E'  B'  can  sustain,  and  it  causes  tension 
in  E'  B'.  Let  the  reader  draw  the  truss 
from  A  up  to  a  section  cutting  E'  C',  E'  B' 
and  D'  B'  and  supply  forces  as  in  fig.  3 
(Art.  9)  equal  to  the  forces  exerted  by  the 
part  of  the  truss  right  of  the  section  upon 
the  cut  members.  Then  since  the  shear 
from  the  left  is  minus  10.5,  it  acts  down, 
therefore  the  supplied  force  acting  on  E' 
B7  at  the  section,  acts  up,  since  its  ver- 
tical component  acts  up,  hence  E'  B'  is  in 
tension.  This  minimum  tension  in  E  B' 
for  an  apex  load  at  C'  is  also  the  mini- 
mum tension  in  B  E  for  an  apex  load  at  C. 

If  a  counter  extended  from  C'  to  D',  it 
could  never  be  in  action  for  the  panel 
loads  given,  as  we  have  just  seen  that  E' 


79 


[§38 


B1  is  always  in  action  for  any  position  of 
the  live  load,  since  the  shear  on  it  is  al- 
ways minus. 

The  rule  therefore  is,  never  insert  a 
counter  in  a  panel  where  the  shear  is  al- 
ways minus. 

A  counter  is  required  in  the  4th  panel 
for  live  load  from  A'  to  E'.  Similarly  a 
counter  is  required  in  the  3d  panel  for  a 
live  load  from  A  to  E  (its  maximum  shear 
being  4.5),  and  no  others  are  needed. 

On  multiplying  the  max.  and  min.  shears 
above,  on  the  inclined  web  members,  by 
sec.  i  •=•  sec.  45°  =  1.414,  their  stresses 
are  obtained. 

The  maximum  and  minimum  stresses  in 
all  the  web  members  are  given  in  the 
following  table  in  pounds,  -f-  indicating 
tension,  —  compression : 


AB 

BC 

B  E 

D  E 

D  G 

E  F 

FG 

Max.    —95445 
Min.    —31815 

-f24000 
-f  600( 

—  G1509  —  25500  +31815 
:_;4S4T  —  30COJ         0 

-f6363 

—7500 
—  3COO 

The  chord  stresses  are  a  maximum  for  a 
full  live  load.    Therefore,  by  formula  (12) 


§  38  ]  80 

Art.  26,  since  P  =  9  +  18  =  27,  d  =  20, 
h  =±  20, 

£n  =  £n  =  n  m  X  13.5 
In  this  formula,  n  is  the  number  of 
panel  lengths  from  A  to  the  centre  of 
moments  for  the  chord-piece  considered, 
and  m  =  6  —  n  is  the  remaining  number 
of  panels.  The  chord  stresses  in  pounds- 
are  as  follows : 

A  C  E,  n  =  1,    67,500 

B  D  or  E  G,  n  =  2,  108,000 

D  F,  n  =  3,  121,500 

The  minimum  chord  stresses  due  to  dead 
load  only  are  -fr  of  the  above. 

Example. — Compute  the  stresses  in  all  the  mem- 
bers of  a  through  Pratt  truss  of  140  ft.  span 
divided  into  7  panels,  and  with  a  height  of  26  ft. 
8  inches,  for  a  dead  load  of  10,500  Ibs.  per  panel 
(of  which  §  is  supposed  to  be  borne  by  lower  apices 
and  J  by  upper  apices)  and  a  live  panel  load  of  21, 
600  pounds.  Use  the  method  of  apex  loads.  The 
dead  panel  load  by  formula  (2),  Art.  5,  is  found  to 
be  10,440  for  a  width  of  bridge  of  24  feet.  The 
live-panel  load  allowing  90  Ibs.  per  sq.  ft.  of  floor 
is  24  X  20  X  90  -*-  2  =  21,600.  The  loads  as- 
sumed are  thus  seen  to  be  usual  ones.  Determine 
where  counters  should  be  used. 


81 


THROUGH  HOWE  TBUSS. 


[§39 


39.  (See  figure  12.)  Take  the  same 
span,  height,  number  of  panels,  dead  and 
live  loads  as  in  the  bridge  first  examined 
in  Art.  38.  The  shears  are  as  there  given. 
Read  the  note  on  counters  in  Art.  25.  As 
the  shear  is  minus  ( —  10.5),  first  in  the 
5th  panel,  no  counter  is  needed  there,  as 
the  main  brace  then  is  in  action  (under  its 
minimum  stress).  Hence  counters  were 
omitted  from  panels  2  and  5,  though  in 
practice  they  are  put  in  wooden  Howe 
bridges  in  every  panel  to  keep  the  main 
braces  in  place,  and  also  as  some  support 
to  them  from  lateral  flexure. 


By  Art.  12  we  have  the  rule:  when  the 
shear  on  a  diagonal  is  plus,  if  the  top  of 
the  diagonal  leans  away  from  the  left 


§39]  82 

abutment  the  diagonal  is  in  compression ; 
if  the  shear  is  minus  and  the  top  leans 
towards  the  left  abutment  the  diagonal  is 
again  in  compression. 

Hence  passing  sections  cutting  each 
diagonal  in  turn;  we  have 

Live  load  from 

A'  to  C,  max.  shear  in  A  B  —  +  67.5 

A'  to  E,     "         «      "  C  D  =  +  43.5 

A'  to  G',    "         "      "  E  F  =    +  22.5 

A'  to  E',    "         "      "  D'  G  =    +    4.5 

A'  to  C',  min.  shear  in  C'  D'  =     -  10.5 

By  the  rule  above,  all  of  these  diagonals 

are  in  compression  for  the  loading  assumed. 

Vertical  B  (7. — Pass   a   section    cutting 

A  C,  B  C  and  B  D.     For  max.  shear,  the 

live  load  extends  from  A'  to  C,  as  it  must 

always   reach  from   A'   to   the   apex   just 

right  of  the  section  (Art.  13).     The  shear 

is  +  67.5  minus  load  at  B  (3)  =  +  64.5, 

hence  supposed  force  at  section  acting  on 

B  C  must  act   down,  .*.  max.  stress  in  tie 

B  C  =  64.5  tension. 

Vertical  D  E. — Section  must  cut  C  E, 
D  E,  D  G  and  D  F.  Is  the  counter  in 
action  ?  With  load  from  A'  to  E,  shear  in 


83  [  §  39 

C  D  is  -f  43.5  and  in  panel  3,  (43.5  —  27), 
a  plus  quantity  .*.  E  F  is  in  action  and 
D  G  out  of  action.  Hence  the  section 
cuts  only  three  members  in  action  and  the 
stress  in  D  E  is  equal  to  the  shear  in  it,  or 
43.5  —  dead  load  at  D  (3)  =  40.5  tension. 
Vertical  F  G. — With  a  live  load  from 
A'  to  G,  the  shear  in  panel  3  is  +  22.5 
and  in  panel  4,  -f-  22.5  —  27  =  —  4.5; 
hence  F  E  and  F  E'  both  suffer  compres- 
sion and  the  counters  D  G  and  D'  G  are 
not  in  action.  Therefore  F  G  holds  up 
only  a  lower  panel  of  dead  load  and  one 
of  live  load ;  hence  max.  tension  in  F  G  = 
6  +  18  =  24. 

F  G  must  also  be  tested  with  a  live  load 
from  A'  to  E',  which  gives  a  shear  in  panel 
4  of  +  4.5  and  a  shear  in  panel  3  of  -j- 
4.5  +  9  =  +  13.5,  for  which  E  F  and 
G  D'  are  in  action  and  D  G  and  F  E'  out 
of  action.  A  section  can  then  be  passed 
through  E  G,  F  G  and  F  D'  (alone  in 
action  on  the  section)  and  the  stress  in 
F  G  is  seen  to  be  equal  to  the  shear  in  E  F 
(+  13.5)  minus  the  load  at  F  (3)  or  10.5; 
or  it  is  equal  to  the  shear  in  G  D'  (-f-  4.5) 


139] 


84 


plus  the  dead  load  at  G  (6)  =  10.5.  This 
is  less  than  the  24  above.  The  minimum 
stress  on  F  G  is  for  dead  load  only.  The 
counters  are  then  out  of  action  and  F  G 
sustains  only  the  dead  load  at  G  —  6. 

For  minimum  stress  on  vertical  D'  E' 
the  live  load  extends  from  A' to  C'.  The 
shear  in  panel  5  =  —  10.5;  in  panel  4,  — 
10.5  +  9  =  -  1.5,  .',  D'  C'  and  F  E'  are 
both  in  action  and  G  D'  out  of  action. 
Hence  a  section  can  be  passed,  as  in  fig. 
13,  cutting  only  three  members  in  action. 
The  shear  in  D'  E'  =  —  10.5  +  3  (load 


atD')  —  —  7.5.  As  it  acts  down,  V  must 
act  up,  giving  a  tension  in  D'  E'  =  7.5. 
This  is  its  minimum  stress,  as  the  live 
load  covers  the  shorter  segment. 

The   minimum   stress   in  D  E  is  conse- 


85 


[§39 


quently   7.5   tension    when  the   live  load 
extends  from  A  to  C. 

It  is  well  to  remark  that,  with  other  panel  loads 
than  the  ones  assumed,  it  could  very  well  happen 
that  for  a  live  load  from  A'  to  C',  the  shear  in 
panel  5  might  be  minus  and  in  panel  4,  plus  ;  in 
which  case  D'  C'  and  D'  G  are  in  action,  whence 
F  E'  is  out  of  action  and  D  E  sustains  only  the 
dead  load  G  at  E',  which  in  this  case  would  be  its 
minimum  tension. 

The  minimum  shear  in  D'  C'  =  —  10.5 
for  live  load  from  A'  to  C' ;  hence  the  min- 
imum compression  in  D  C  is  10.5  sec  i  for 
a  live  load  extending  from  A  to  C.  The 
minimum  tension  in  B  C  =  19.5  and  the 
minimum  compression  in  A  B  is  due  to 
a  shear  22.5,  both  for  dead  load  alone. 

The  maximum  and  minimum  stresses  in 
pounds  on  the  web  members  are  then  as 
follows : 


AB 

BC 

CD 

DE 

EF 

DG 

FG 

MAX. 
MIN. 

—95445 
—31815 

+64500 
+19500 

—61509 
—14847 

+40500 
+  7500 

—31815 
0 

—  6363  j+24000 
0  U  6000 

The  chord  stresses  are  found  as  in  the 
preceding  article  from  the  formula, 


§  40  ]  86 

tn  =  cn  —  n  m  X  13.5 ; 

or  expressing  them  in  pounds,  and  remem- 
bering that  n  is  the  number  of  panel 
lengths  from  A  to  the  center  of  moments 
for  the  chord  piece  considered,  we  have 
the  maximum  stresses: 

A  C  or  B  D,  n  =  1,  .  .  .  67,500  Ibs. 

C  E  or  D  F,  n  =  2,  .  .  ,  108,000  Ibs. 
E  G,  n  =  3,  .  .  .  121,500  Ibs. 

The  minimum  stresses  due  to  dead  load 
only  are  A  of  these. 

Example, — Consider  a  through  Howe  truss  of 
the  dimensions  and  loading  given  in  the  example 
at  the  end  of  Art  38.  Ascertain  where  counters 
should  be  used  and  the  stresses  in  all  the  members. 

DECK  PKATT  TKUSS. 

40.  Let  figure  14  represent  a  Pratt  truss 
of  126  feet  span,  divided  into  7  panels  each 
18  feet  in  length.  To  ensure  counters  in 
the  three  central  panels,  a  large  live  load 
of  2,000  Ibs.  per  ft.  on  one  truss  will  be  as- 
sumed to  pass  over  the  upper  chord.  Ex- 
pressed in  thousand  pounds,  this  gives  a 
panel  load  of  p  =  2  X  18  =  36.  Apex 


87  §  40  ] 

loads  of  6  and  3  will  be  assumed  at  the 
upper  and  lower  chords  respectively  to 
represent  the  dead  load. 

An  illustration  will  now  be  given  of  the 
working  of  "  the  exact  method  "  for  shears, 
Art.  33;  this  will  not  be  found  quite  so 
simple  as  the  method  of  "  apex  loads." 

Equation  (15)  reduces  in  this  case  to  S, 
:=:  3  (7  —  ft)2;  and  it  must  be  remembered 
that  the  live  load  extends  from  A'  to  the 
apex  just  right  of  the  section  taken,  a  dis- 
tance, c  =  md  and  in  addition,  x  = 


N— 1 


=  i  c  to  the  left  of  the  apex. 


The  dead-load  reaction  =  3  X  9  —  ^7, 
from  which  we  find  dead-load  shears  as 
usual.  On  adding  them  to  the  live-load 
shears  computed  from  the  formula  above, 


2 

75  +  18 

=  +    93 

3 

48+9 

-+57 

4 

27+0 

=  +    27 

5 

12-9 

=  +      3 

6 

3  —  18 

=  —    15 

7 

0  —  27 

-  —    27 

§  40  ]  88 

we  get  the  totals  given  below  for  the  di- 
agonals: 

c  =  A'B,n=I  S  =  108  +  27  =  +  135 
c  =  A'D 
c  =  A'F 

c  =  A'  D' 
c  =  A  B' 

c—     0 

The  maximum  shears  in  the  diagonals 
A  C,  B  E,  D  G,  F  G7  and  F'  E'  are  given 
by  the  values  of  S  above  for  n  =  1,  2,  . . . ,  5. 
As  the  shear  is  minus  in  the  sixth  panel, 
no  counter  is  required  there  or  in  the  2d 
panel;  but  the  three  central  panels  must 
have  counters. 

For  maximum  stress  in  a  post,  as  F  G, 
the  load  is  placed  as  drawn  in  fig.  14,  and 
the  section  passed  through  the  five  pieces 
as  shown.  To  test  what  pieces  are  in 
action  we  must  know,  where  counters  are 
cut  by  the  section,  the  amount  of  live  load 
held  up  at  the  apex  just  right  of  the  sec- 
tion. Thus,  for  a  load  2  thousand  pounds 
per  foot,  F  in  the  figure  holds  up  part  of 


89  [§40 

the  load  2x  to  left  of  F  and  a  half-panel  load 
from  the  right,  or 

18  +  2  x  —  *j£  =  18  +  6  m  —  %  m\ 
2  d 

since  x  =  e  c  =  e  m  d. 

From  this  formula  we  find  for 
C  =  A'  D,  live  load  held  at  D  =  35.5 
C  =  A'  F,     "      "       "      "  F  =  34.0 
C  =  A'F,    "      "       "      "  F'  =  31.5 
C  =  A'D',  "      "       "      "D'  =  28.0 
For  brevity,  the  shear  on  a  diagonal  in 
panel  4  for  a  certain  position  of  the  load 
will  be  called  S4,  and/br  the  same  position 
of    the   load  the   shears   on   diagonal   in 
panels  3  and  5  will  be  called  S3  and  S5  re- 
spectively.   Similarly  for  any  panel.    Also 
observe  that  the  shear   on   F   G,   at  the 
section,  fig.  14,  when  E  F  and  G  F'  are 
out  of  action,  is  equal  to  the  left  reaction 
minus  all  the  loads  up   to  the  section,  or, 
in  this  case,  to  S3  —  load  at  G  =.  57  — 
3  =  54. 

Bearing  in  mind  the  rules  of  Art.  12  and 
what  precedes,  the  reader  should  find  no 
difficulty  in  following  the  brief  consider- 
ation of  posts  that  follows. 


§40]  90 

Post  B  C;  c  —  A'  B;  max.  stress  in 
B  C  =  Si  —  3  =  135  —  3  =  132. 

Post  D  E  ;  c  =  A'  D;  S2  =  93,  S3  =  93 

-  (35.5  +  9)  or  plus,  /.  D  G  is  in  action 
and  E  F  not  in  action,  .'.   max.  stress  in 
D  E  =  93  —  3  =  90. 

Post  F  G  ;  c  --  A'  F;  S3  ==  57,  /.  E  F 
out  of  action.  S4  =  57  —  (34  +  9)  or 
plus  .'.  G  F7  is  out  of  action  and  the  sec- 
tion cuts  only  three  pieces  in  action,  .'. 
maximum  stress  in  FG  =  =  Ss  —  3  —  57 

-  3  =  54. 

Post  F'  G'. — The  min.  stress  is  for  dead 
load  only,  when  none  of  the  (dotted)  coun- 
ters act,  and  is  equal  to  dead  load  at  F'  : 
6.     The  same  holds  for  F  G. 

Post  D'  E'.  For  min.  stress,  try  c  —  A' 
B'  and  pass  a  section  through  G'  E',  F'  E', 
D'  E'  and  D'  B'.  S6  =  -  15;  to  find  S5 
we  first  find  the  live  load  held  up  at  D'  - 

wx*         2  X    32 


2  d        2  X  18 

•'•     S5     ~     S6    +  10ad8   at  I)/   aild  E/  =  ~ 

15  _f_  9  _j_  .5  —    _  5.5^  /.  G'  D'  is  in  action 
and  F'  E'  out  of  action;  hence  the  section 


91  [  §  40 

cuts  but  three  members  in  action  and  the 
live  load  covers  the  smaller  segment  of 
the  span  .'.  the  shear  in  Post  D'  E'  is 
a  minimum  and  is  equal  to  Sc  -f-  load  at 
E'  =  -15  +  3=  -12. 

[For  other  loads  than  those  given,  it 
might  happen  that  F'  E'  as  well  as  E'  B' 
was  in  action ;  in  this  case  E'  D'  supports 
only  the  loads  at  its  top,  or  6  +  amount  of 
live  load  held  at  D'.] 

The  minimum  stress  in  post  D  E  when 
c  —  A  B  is  thus,  for  the  loads  first  taken  — 
12.  For  G  =  A'  B',  we  have  smallest 
shear  in  E7  B7  =  S6  =  —  15;  this  is  also 
equal  to  the  smallest  shear  in  B  E  for  e  = 
AB. 

Under  dead  load  only,  A  C  receives  a 
shear  of  27  and  B  C  of  27  —  3  =  24.  The 
diagonals  in  all  counter-braced  panels  have 
a  minimum  stress  zero,  for  when  one  is 
in  action  the  other  is  out  of  action,  and 
vice  versa. 

Example. — Compute  the  maximum  and  mini- 
mum shears  in  the  web  members  of  the  truss 
just  considered,  by  the  method  of  apex  loads,  and 
compare  the  results  with  those  just  obtained. 


§41]  92 

41.  Deck  Howe  Truss. — The  investiga- 
tion of  the  trusses  previously  given  should 
enable  the  attentive  student  to  place  the 
loads  for  maximum  and  minimum  shears, 
and  draw  the  proper  sections  for  a  deck  Howe 
truss.  Assume  the  truss  of  8  panels  with  a 
dead  apex  load  on  the  upper  chord  of  7,  and 
on  the  lower  chord  of  3,  the  live  apex  load  on 
tipper  chord  being  16;  all  loads  being  in 
thousand  pounds.  Letter  the  truss  as  in 
fig.  12,  Art  39,  calling  the  centre  vertical 
H  I.  For  a  deck  truss,  vertical  posts  should 
be  placed  at  A  and  A',  fig.  12,  to  meet 
the  upper  chord  produced.  The  additional 
pieces  are  not  parts  of  the  truss  proper. 
Use  the  approximate  method  of  apex  loads, 
Art.  32,  and  attend  strictly  to  the  principles 
of  arts.  12  and  13.  The  sections  taken 
must  never  cut  more  than  three  pieces  in 
action. 

The  maximum  and  minimum  shears  for 
the  braces  are  as  follows: 

AB  CD  E  F  GH  FI 

Max.      91  67          45  25  7 

Min.       35  23  9  0  0 


93  [  §  42 

The  stresses  for  the  vertical  ties  are  : 

BC  DE  FG  HI 

Max.      70  43  28  10 

Min.      26  12  33 

For  d  —  20,  h  =  26,  we  have  the  fol- 
lowing maximum  chord  stresses: 
A  C,  B  D,  stress  =  70, 
C  E,  D  F,  stress  =  120, 
E  G,  F  H,  stress  =  150, 
G  I,  stress  =  160. 

42.  Remark  on  Counters. — When  a 
Pratt  bridge  with  counters  is  first  swung 
clear  of  false- works,  the  main  ties,  but  not 
the  counters,  are  in  action.  If  a  light  load 
is  put  on,  and  the  counters  made  taut;  for 
a  heavier  or  full  load  they  will  be  loose 
and  for  the  dead  load  only,  under  slight 
initial  stress.  This  is  because  the  loading 
causes  increased  deflection  of  the  truss,  in- 
volving lengthening  of  main  ties  and  short- 
ening of  counters. 

If  an  initial  stress  is  put  in  the  counters, 
with  no  live  load  on  bridge,  the  main  ties 
have  an  increase  of  tension  equal  to  this 
stress.  Thus  in  fig.  14,  Art.  40,  we  suppose 


§  43  ]  94 

an  initial  stress  in  counter  E  F,  whose 
vertical  component  is  V,  and  conceive 
the  pieces  E  G,  D  G,  E  F  and  D  F  severed 
by  a  plane.  On  supplying  forces  at  the 
cut  parts  equal  to  the  action  of  the  right 
part  of  the  truss  upon  the  cut  members,  we 
have,  as  in  arts.  9  — 11,  R  —  2  w  -f-  Y7  = 
V,  since  Y7  acts  up  and  V  down;  or  the 
vertical  component  Y,  of  the  stress  in  D  G 
has  been  increased  by  V7 ;  hence  the  stress 
in  D  G  has  been  increased  by  the  initial 
stress  (Y7  sec  i)  in  the  counter  E  F. 

If  the  initial  stress  in  E  F,  and  the  in- 
crease due  to  it  in  D  G,  are  resolved  into 
vertical  and  horizontal  components  at  D, 
E,  F  and  G,  they  balance  along  posts  and 
chords,  and  do  not  affect  any  other  panel. 

THE  WHIPPLE  TRUSS. 
43.  For  long  spans,  over  180  feet  say, 
the  inclination  of  the  diagonals,  for  usual 
panel  lengths  and  heights,  of  the  trusses 
previously  examined,  is  not  the  best  for 
economy.  Hence,  the  use  of  multiple  sys- 
tems having  the  ties  cross  two  or  more 
panels,  as  in  the  once  popular  Whipple  truss, 
fig.  15. 


95 


[§43 


In  the  truss  of  12  panels  shown,  we 
shall  first  consider  the  dead  load  per  panel 
w,  as  concentrated  at  the  lower  apices,  and 
afterwards  regard  it  as  divided  in  the 
usual  proportion  between  the  two  chords. 
The  loads  w  are  represented  by  circles 
placed  below  the  chord  and  the  live  load 
per  panel  p9  by  rectangles  placed  just 
above  the  lower  apices.  It  will  be  noticed 
that  any  weight,  as  that  at  /",  can  travel 
to  either  r.batment  by  only  one  web  system, 
as  a  B  d  D  /  H  h  J  j  L  m.  The  weight  at 
e  must  follow  the  other  system,  a  B  c  C  e 
G  g  I  i  K  k  L  m.  If  the  counter  (dotted 
line)  is  not  in  action,  then  the  main  tie 
replaces  it,  and  vice  versa.  We  observe, 

BCDEFGH1J.KL 


Fig.  15. 

therefore,  that  the  black  weights  travel 
towards  either  abutment  only  by  the  first 
system ;  the  others,  by  the  second. 

The  loads  at  b  and  I  are  carried  to  B  and 


§  43  ]  96 

L  by  the  vertical  suspenders  Bfi  and  L/, 
and  they  will  be  regarded  as  divided 
equally  between  the  two  systems.  For 
the  chord  stresses  the  results  are  the  same, 
whatever  division  of  these  loads  at  b  and  I 
we  make  between  the  two  systems. 

For  the  shears,  if  the  whole  of  p  at  I  be- 
longs to  one  system,  the  error  in  the  left 
reaction  (and  therefore  the  shear)  over  the 

P 

above    supposition    is    only    T^  ^,  a  very 

small  amount.  With  vertical  end  posts, 
as  in  fig.  16,  there  is  no  uncertainty,  as 
then  any  load  at  b  or  I  can  travel  by  one 
system  only. 

It  may  occur  to  the  reader  that  this 
truss  has  superfluous  members.  Leaving 
out  the  counters,  as  they  do  not  act  when 
the  main  ties  are  in  action,  m  =  number  of 
pieces  —  45,  n  =  number  of  joints  —  24 
/.  m  —  2  n  —  3,  and  the  test  of  Art.  2  is 
fulfilled ;  hence  the  figure  of  the  truss  has 
no  superfluous  lines,  but  just  enough  sides 
to  completely  fix  its  form. 

In  finding  stresses,  the  method  used  is 
to  consider  the  partial  truss  along  which 


9V  [  §  43 

the  black  weights  travel  as  acting  inde- 
pendently of  the  other  partial  trass,  and 
properly  combine  the  results  where  the 
two  trusses  have  the  same  member  in 
common. 

Thus,  to  find  the  shears  in  the  diagonals 
and  end  posts  for  the  dead  loads,  let  us 
first  consider  the  partial  truss  a  Be..., 
with  loads  w  at  c,  e,  g>  i,  Jc  and  i  w  at  b 
and  Z;  the  left  reaction  is  thus  i  6  w  =  3  w- 
Hence  shear  in  B  c  =  3  w  —  £  w  (at  b)  = 
2i  w\  in  C  e  =  2i  w  —  w  (at  c)  =  li  w; 

in  E  g  =  -~ ;  in  g  I  or  G  i  (whichever  is  in 

/v 

action  when  live  load  is  added)  =  %  w  — 
w  (at  g)  =  — -  i  w,  and  so  on. 

For  the  black  weight  system  we  have 
similarly,  reaction  =  2%  w\  shear  in  B<# 
=  2  w ;  in  D  f  =  1  w ;  in  F  h  or/"H  =  o ; 
in  Hj  or  H  J  =  o  —  w  =  —  w,  and  so  on. 

Notice  that  these  shears  are  the  same 
for  the  inclined  web  members,  whether 
the  dead  load  is  supposed  concentrated  at 
the  lower  chord,  as  above,  or  divided  be- 
tween the  two  chords  in  any  manner. 

The  live- load  shear  on  a  tie,  when  the 


§43] 


98 


load  extends  from  in  to  its  foot,  is  equal  to 
the  left  reaction.  Thus,  when  the  load  is 
placed  as  in  fig.  15,  we  get  the  maximum 
shear  on  13  d  due  to  the  weights  p  at  </,/', 
A,  j  and  I  p  at  /,  equal  to 

(9  +  ,  +  5#s+o.<0  =  -^,. 


Similarly  we  proceed  for  any  of  the 
other  ties  or  counters.  The  shears  are  as 
follows: 


Live-load 
from 

Member. 

Dead-load 
Sbear 

Live-load 
Shear 

m  to  b 

aB 

+  5.5  W 

+  5.5  p 

"  "  c 

B  c 

+  2.5  10 

_j_   :?o.r,     ^ 

"  "  d 

Br7 

+  2.0  t0 

I       24"f)      n 

,l  '-'. 

"    "    6 

C  e 

+  1.5  w 

la"    ^ 

«   «y 

~Df 

+  1.0  ?o 

_|_     15.5     ^ 

«  «^ 

Eff 

+  C.o  w 

i       li  ;>       p 

r:       i 

"  "  A 

Tfh 

—  0.0  w 

+     *  '     f> 
12     -^ 

U     ((     £ 

G  i 

—  0.5  w 

_|_      6ft      /^ 

12     *^ 

"  "  J 

H./ 

-  1.0  w 

+      -      /> 

"  "  k 

t  K 

—  1.5  w 

+    TJ    ^ 

«  «  ^ 

/  L 

—  2.0  w 

12     •* 

"  «  ^ 

^  L 

—  2.5  w 

T2     -^ 

99  [ §  43 

We  have  assumed  here  that  counters 
would  only  be  needed  where  placed  in  the 
figure;  hence  the  sum  of  the  shears  above 
for  any  member  gives  its  greatest  total 
shear  except  for  the  last  three  members, 
where  the  sum  is  supposed  to  be  negative, 
and  the  total  shear  a  minimum,  as  the 
shortest  segment  of  the  span  is  covered 
by  the  live  load. 

Of  course  we  multiply  each  total  shear 
by  the  secant  of  the  angle  the  web  piece 
makes  with  the  vertical  to  get  the  corre- 
sponding stress. 

The  counters  and  the  main  ties,  crossing 
at  their  centres,  have  zero  for  their  mini- 
mum stress.  The  minimum  stress  on  a  B 
and  b  B  is  due  to  dead  load. 

Let  us  consider  next  the  vertical  posts. 
They  receive  the  vertical  components  (~ 
shear  in  amount)  of  the  stress  in  the 
diagonals  connecting  with  their  tops,  and 
in  addition  any  part  of  the  dead  load 
supposed  to  be  concentrated  there.  Their 
maximum  or  minimum  stresses  are  thus  the 
maximum  or  minimum  shears  (irrespective 
of  sign)  in  the  ties  or  counters  connecting 


§  43  ]  100 

with  their  tops  increased  by  the  dead-apex 
load  on  the  top  chord. 

When  the  counters  are  out  of  action 
that  meet  at  F,  G  and  H  (as  they  are  for  a 
full  live  load)  the  posts  F/,  G  g  and  H  h 
sustain  only  the  deads  loads  at  their  tops, 
their  minimum  stresses.  For  such  loading 
as  causes  counter  G  i  to  act,  g  I  is  out  of 
action,  and  post  I  i  sustains  only  the  dead 
load  at  I.  Similarly  the  minimum  stress 
in  post  J  j  is  the  dead  load  at  J,  HJ  being 
in  action  and  h  J  out  of  action.  Similarly 
for  E  e  and  T>  d.  The  posts  K  k  and  C  c 

bear  the  minimum  stress  (1.5  w  —  -  -  p)  -f- 

1/c 

dead  load  at  top,  as  explained  above. 

The  hypothesis  that  the  two  partial 
trusses  act  independently,  forms  the  basis 
for  the  investigation  of  any  compound  or 
"  double-intersection  "  system.  This  is  not 
actually  true,  as  the  chords  of  the  two  sys- 
tems cannot  change  length  independently 
under  loading;  but  it  is  probably  suffi- 
ciently near  for  practice.  For  railroad 
bridges  where  actual  wheel  loads  are  used, 
the  computation,  even  on  this  inevact  hy- 


pothesis,  is  very  laborious ;  so  that,  for  both 
reasons,  specifications  now  give  the  prefer- 
ence to  those  trusses  in  which  the  stresses 
can  be  accurately  computed. 

The  maximum  stresses  in  the  chords  are 
for  a  full  live  load.  Call  w'  =  w  +  p\ 
find  the  shears  on  each  web  member  due 
to  w1  at  each  lower  apex,  and  apply  the  rule 
of  Art.  22.  It  is  well  to  mark  the  shears 
on  the  members,  as  is  done  in  fig.  16,  for  a 
Whipple  truss  with  vertical  end  posts.  On 
decomposing  the  stresses  in  the  diagonals 
at  each  end,  as  shown,  into  vertical  and 
horizontal  components,  the  latter  give  the 
chord  increments.  Thus,  in  fig.  16,  stress 

in  A  B  =  3  w1  ^  +  2.5  w1  ~  =  8  w1  r 
h  h  h 

2  d 

Stress  in  B  C  =  stress  in  A  B  -f  2  w1  — - 

h 

10     id 

—   12  w'  — 

h 
Similarly  at  the  apices  of  either  chord. 

[Notice  that  the  chord  stress  is  a  maxi- 
mum where  the  shear  is  zero.] 

For  the  truss  fig.  15,  a  similar  method 


102 


Zw'tvml 

A  2jw'  B  2 


Fig.  16. 

of  chord  increments  (Art.  21)  applies,  giv 
ing  the  stress  in 


a  b  c 


/ 

•=.     O.D  X  w'  — 


—  . 
h 


cd  =    8. 

BC,  r76  -  12 
C  D,  ef  =  15 
D  E,f#  =  17 
E  F,  F  G  =  18 


X 
X 
X 
X 
X 


The  last  result  can  be  tested  by  taking 
moments  about  g,  since  the  counters  are 
out  of  action.  The  reaction  for  the  fully 
loaded  truss  is  5.5  w'.  There  are  full 
loads  at  b,  c,  d,  e  and  /*,  whose  resultant 
acts  at  d.  Hence  stress  in  E  I  equals  (5.5 
to/  x  6  d  —  5  w1  X  3  d)  -4-  h  =  18  w1 
d  -T-  h,  as  found  above. 


103  [  §  44 

When  formulas  have  been  found,  as  in 
this  case,  applicable  to  anyWhipple  truss  of 
12  panels,  it  is  best,  for  given  values  of  ic, 
jt>,  d  and  h,  to  compute  the  successive  re- 
sults by  evident  successive  subtractions  for 
shears  and  additions  for  chord  stresses,  and 
check  the  results  in  the  end  by  the  formula. 

Example. — Deduce  formulas  similar  to  those 
above  for  shears  and  chord  stresses  for  a  through 
Whipple  truss  of  1 1  panels,  supposing  only  four 
counters  needed,  those  meeting  at  F  and  G  (fig. 
15) ;  also  give  maximum  and  mimimum  stresses 
in  the  posts.  Counters  are  not  needed  for  those 
panels  where  the  total  shear  is  minus  (Art.  38);  but 
this  can  be  ascertained  only  when  numerical  values 
have  been  inserted  in  the  formulas.  In  all  trusses 
counters  are  frequently  inserted  where  not  theo- 
retically needed  for  the  loads  assumed,  as  still 
heavier  live  loads  might  necessitate  their  use. 

LATTICE  TRUSS. 

44.  The  seven,  panel  lattice  truss,  fig. 
17,  can  be  divided  into  two  systems;  the 
one  shown  by  full  lines  and  the  other  by 
the  dotted  lines,  called  respectively  the 
full  and  the  dotted  systems.  The  loads 
(live  or  dead)  held  at  B,  C  and  B'  C7, 


§44]  104 

will  be  divided,  as  before,  equally  between 
the  two  systems.  Call  p  the  panel  live-load, 
then  the  max.  live-load  shear  in  A  B  =  i 
6  p  =  3  p.  The  oilier  maximum  live-load 
shears  are  equal  to  the  left  reactions  for 
the  full  or  the  dotted  systems  for  loads 
covering  the  longer  segment  of  the  span, 
and  are  as  follows: 

B  E ;  p  at  E,  G'  and  0.5  p  at  C;, 
shear  =  p  (f  +  f)  -f  \p  X  \  =  +  1.216  p 

CDandDG;^  at  G,  E7  and  0.5  p  at  C', 
shear  =  #(t  +  |)+i^X-f=  +  0.929  p 

E  F  and  F  G7 ;  p  at  G'  and  \  p  at  C', 
shear  =  p  x  f  +  3  ^  X  |  =  +  0.5  j?. 

For  minimum  shears  on  these  same 
pieces,  it  will  be  simplest  here  to  regard 
the  live  load  as  extending  from  the  left 
abutment  to  the  nearest  apex  left  of  the 
section.  The  left  reaction,  minus  the  loads 
up  to  the  section,  gives  the  shear 

BE;  i^atC, 

shear  =  f  =  — f  =  —  .072  p. 

/£  7        /w 

CD  and  D  G;  %p  at  C,  shear  ==  -.072 .p. 
E  F  and  F  G7;  \p  at  C  and  p  at  E, 


105 


[§44 


Shear  = 
.3571?. 


X  f +.P  Xf--  1.5j?  = 


On  multiplying  each  of  these  shears  by 
sec  A  B  C  =  s  and  attending  to  rules  of 
Art.  12,  we  find  the  stresses  in  the  table 
below,  corresponding  to  live  load. 

For  the  dead-load  shears,  -§-  w  will  be 
assumed  as  the  dead  load  held  at  each 
lower  apex  and  ^  w  at  each  upper  apex,  w 
being  the  dead-panel  load.  The  loads  at 
B,  C  and  B',  C',  on  either  the  full  or 
the  dotted  systems,  being  symmetrically 
placed  with  respect  to  the  centre,  give 
reactions  equal  \  w  on  each  system,  to 
which  must  be  added  reactions  caused  by 
the  remaining  loads.  For  the  full  system, 
we  have  the  remaining  loads  -^wat  F  and 
D',  f  w  at  E  and  G' ;  therefore,  the  left  re- 
action R  equals 


§44]  106 

i  w  +  4-  [^  w  (4  +  2)  +  f  ^  (5  +  3)  ]  = 
1.548  w. 

For   the   dotted  system,   the    remaining 
loads  are  i  w  at  D  and  F',  f  w  at  G  and 
E7;  hence  the  left  reaction  Rx  equals 
i  10  +  |  [i-  w  (5  -f  3)  +  f  w  (4  +  2)]  = 
1.452  «?. 

The  shear  in  A  B  is  the  sum  of  R  and  R', 
or  3  w,  which  is  otherwise  evident. 

The  shear  in  B  E  =  R  —  loads  on  full 
system  held  at  B  and  C  (.5  w)  =  (1.548 
_  .5)  w  =  1.048  ic.  Similarly,  shear  on 
C  D  =  (1.452  —  .5)  w  =  0.952  w.  Shear 
on  E  F  =  shear  on  B  E  —  load  at  E  — 
(1.048  —  .667)  w  =  0.381  w. 

Similarly,  shear  on  D  G  =  (.952  —  .333) 
w  =  0.619  w;    shear  on  F  G'  =   (.381  - 
.333)  w  —  0.048  w  and  shear  on  G  F7  = 
(.619  —  .667)  w  =  —  0.048  w. 

It  is  best,  in  order  to  avoid  mistakes,  to 
mark  the  stresses  caused  by  these  shears  at 
once  on  a  sufficiently  large  sketch  of  the 
truss,  L  for  live  load,  D  for  dead  load; 
also  the  character  of  the  stress  they  give 
on  each  diagonal.  Notice  that  D  G  and 
G  F7  are  both  in  tension  under  dead  load 


107  [§44 

only,  so  that  the  chord  increment  at  G  is 
(.619—  .048)  w  =  0.571  w. 

The  dead-load  stress  in  B  C  equals  the 
vertical  component  of  the  compression  in 
C  D  plus  the  entire  load  held  at  C  or 
(.952  +  .667)  w  =  1.619  w,  as  B  C  belongs 
to  both  systems. 

The  chord  stresses  due  to  dead  load  are 
quickly  found,  by  the  rule  of  Art.  22,  to 
be  for 

A  C  =  3.000  w  t,  B  D  =  4.048  w  t 
C  E  =  3.952  w  t,  D  F  =  5.619  w  t 
E  G  =  5.381  w  t,  F  F7  —  6.048  w  t 
GG'  =  5.952  wt, 
noting  the  peculiarity  above  at  G. 

The  tangent  of  the  inclination  of  a 
diagonal  to  the  vertical  —  t  =  d  -T-  h. 

Finally,  we  have  to  find  the  chord 
stresses  due  to  a  full  live  load. 

We  proceed,  as  for  dead  load,  in  first 
getting  reactions;  then  shears  and  finally 
chord  stresses,  for  p  at  each  lower  apex,  i 
V  at  C  and  C7  going  to  either  system.  The 
left  reaction  due  to  i  p  at  C  and  i  p  at  C7 
is  i  p ;  therefore,  left  reaction  for  full  sys- 
tem equals 


§44]  108 

\P  +P  (f  +  f)  =  1-643  JP, 
and  for  dotted  system 


The  sum  3  p  is  the  shear  in  A  B.  The 
shears  in  other  members  are  then  as  fol- 
lows: B  C  =  (1.357  +  0.500)  p  =  1.857^; 
C  D  =  (1.357  —  .500) .p  =  .857 p  =  D  G; 
BE—  (1.643  —  .500)  p  =  1.143  p;  E  F, 
F  G'  —  (1.143  i?  —  p)  =  0.143  i?;  F  G 
=  (0.857  p  --  p)  -  -  0.143  p,  giving 
tension  in  F7  G;  also  there  is  tension  in 
D  G,  /,  chord  increment  at  G  =z  (.857 
—  .143)  p  =  .7 14  p.  The  resulting  stresses 
are  given  in  the  following  table: 

In   this   table   -f    indicates   tension,   - 
compression.      When  p,  w,  s,  t  are    given 
the  maximum  and  minimum  stresses  in  all 
the  members  can  be  found  by  combining 
results  above  for  live  and  dead  loads. 

The  diagonals  requiring  counter-bracing 
are  those  whose  maxima  are  of  a  different 
kind  of  stress  from  their  minima  stresses. 

The  truss  just  examined  is  used  for 
riveted  structures  of  75  to  125  ft.  spans. 
With  vertical  suspenders  dropped  from  the 


109 


Live  Loads 
at 

Member 

Stress  due 
to  Live  Load 

Stress  due 
to  Dead  Load 

CtoC' 

AB 

—  3.000J9  8 

-  3.000  W  8 

E,  G  ,  C 

B  E 

+  1.215  ps 

+  1.048  w  8 

C 

BE 

—    .072  p« 

+  1.048WI 

G,  E',  C' 

CD 

—  0.929  .p* 

—  0.952  w  s 

C 

CD 

+  0.072^?* 

—  0.952  w  8 

G,  E  ,  C 

DG 

+  0.929  p  s 

-|-0.619io0 

C 

DG 

—  0.072  ps 

+  0.619  w  8 

G,  C 

EF 

—  0.500  ps 

—  0.381  w  8 

C,  E 

EF 

+  0.357ps 

-  0.381  ws 

G  ,  C 

F  G' 

+  0.500  ps 

-\-  0.048  w  8 

C,  E 

FG' 

-0.357  ps 

+  0.048  ws 

CtoC' 

AC 

+  3.000;?  t 

+  3.000  w  t 

Ci 

CE 

+  3.857  pt 

+  3.952  w  t 

" 

EG 

+  5.143  pt 

+  5.381  w  t 

tt 

GG' 

+  5.857  pt 

+  5.952  w  t 

n 

BD 

-  4.143  p  t 

—  4.048  w  t 

t  < 

DF 

-  5.857  pt 

-  5.619  w  t 

« 

FF' 

-6.143;?$ 

—  6.048  w  t 

(1 

BC 

+  1.857;? 

+  1.619  w 

intersection  of  diagonals  to  lower  chord, 
thus  dividing  the  truss  into  double  the 
number  of  panels,  and  modifications  at 
the  ends,  it  has  been  used  as  a  pin-con- 


§  45  ]  110 

nected  structure  for  the  Memphis  and  other 
bridges  of  very  large  span. 

A  somewhat  similar  method  of  subdivid- 
ing the  panels  of  a  Pratt  truss  gives  us  the 
Baltimore  truss.  When  the  upper  chord 
is  inclined  in  addition,  there  results  the 
Petit  truss,  which  is  the  standard  now  for 
very  long  spans.  The  last  two  trusses 
named  are  really  single  intersection  sys- 
tems, with  secondary  trusses  used  to  sub- 
divide the  panels.  They  thus  admit  of 
ready  calculation  without  the  approxima- 
tions incident  to  double-intersection  sys- 
tems, as  the  Whipple  and  Lattice  trusses 
given  above.  Larger  treatises  must  be 
consulted  for  the  computation  of  the  stresses 
in  such  trusses. 

ANALYSIS  OF  TRUSSES  FOB  WHEEL- LOADS. 
45.  In  the  computation  of  the  stresses 
in  railroad  bridges,  the  standard  load  is 
generally  two  consolidation  engines,  fol- 
lowed by  a  uniform  car-load.  lu  fig.  18 
is  shown  the  wheel  diagram  for  Cooper's 
class  E  40,  taken  from  his  specifications 
for  1896.  The  weights  are  those  on  one 


Ill 


[§45 


rail,    or   for  a  single-track  bridge,    for  a 
single  truss. 


re  -r-^ 
804-30--  M 

—  230 -  —  50-  -  « 

__70-- 

—  #30  -  —  00-  -  o, 

—  1,640  --  103-  -  <* 

—  2,155 -~H6-\-  vi 

—  2,951  --/£ 

--#2-1-  vo 

752-- 


—  5,848  - -172--' 

—  6,708-- 192-  -  \ 


—  8,728  --232-- 

-  10,816  --245--  i 

-  12,041  --258 — ; 

-  13,589-- 271 --* 

'4,944  \-284-\-\ 
L  75,364 


109 


101 


91 


86 


77 


72 


66 


61 


53 


45 


40 


•55 


-50 


21 


16 


10 


8 


ie 


23 


32 


37 


43 


64 


88 


104 


109 


The  figures  aboye  the  wheels  are,  in  order: 
The  weights  on  wheels. 
The  distances  apart  of  the  wheels  in  feet. 
The  distances  from  p  in  feet. 
The  distances  from  wheel  I  in  feet. 


§45]  112 

The  figures  below  the  wheels  are : 

The  number  of  the  wheel. 

The  sum  of  the  loads  from  the  left. 

The  sum  of  the  moments  about  any 
wheel)  of  all  wheel-loads  on  its  left. 

The  uniform  car-load  beginning  at  p  is 
2  thousand  pounds  per  foot;  all  weights 
are  in  thousand  pounds,  and  the  moments 
are  in  thousand-pound  feet. 

The  moments  are  easily  computed.  Call 
the  wheel-loads  wl9  w^  . . . ,  beginning  at 
the  left,  and  designate  the  sum  (wl  +  w^ 
+  . . .  -f  wn)  by  ^_n ;  also  let  Mn  represent 
the  sum  of  the  moments  of  w\_n  about 
wheel  n  and  Mn+1  the  sum  of  the  moments 
of  W!_(n+i)  about  wheel  n  -f  1. 

If  the  resultant  of  the  loads  wl  _  n  acts 
at  a  distance,  c  to  the  left  of  wheel  n,  then 
Mn  =  wl  _  n  c,  and  calling  a  the  distance 
between  wheels  n  and  n  +  1,  Mn  + 1  — 
w  a  _  n  (c  +  a)  =  Mn  +  wt  _  „  a.  Thus 
since  M2  —  10  X  8  —  80  and  wheels  2 
and  3  are  5  feet  apart,  M3  =  M2  +  wl  _  2 
X  5  =  80  +  (10  +  20)  5  ==  230;  M4  = 
230  +  50  X  5  =  480,  and  so  on. 


113  [§46 

46.  Reaction  for  Wheel-loads.  —  It  is 
frequently  necessary  to  find  the  left  reac- 
tion R  when  only  the  wheel-loads  (wl  +  w9 
+  111+  wn)  are  on  the  bridge  ;  call  the  dis- 
tance of  wheel  n  (which  is  nearest  the  right 
abutment)  from  the  right  abutment  e  ;  then 
the  moment  about  that  abutment  of  wl  „  n 
by  the  above  =  w1  _  n  (c  -f-  0)  =  Mn  -f  wv  _ 
e  ;  and  if  the  span  is  I  feet, 


.(18) 


Thus,  if  for  a  span  of  100  feet,  wheel  14  is 
5  ft.  to  left  of  right  abutment,  R  =  (87*8  -f- 
232  X  5)  -T-  100  =  75.68,  the  quantities 
8728  and  232  being  found  from  the  diagram 
under  14.  Sometimes  R  is  required  when 
wheels  1  and  2  (or  more)  are  off  the  span. 
In  this  case,  as  Mn  and  n\  _n  include  these 
loads,  we  must  subtract  the  sum  of  their 
moments  about  wheel  n  from  Mu,  and  the 
sum  of  the  loads  off  the  span  from  wl  _  n  be- 
fore applying  the  formula. 

Thus  for  the  100-feet  span,  let  «018  be  5 
feet  to  left  of  right  abutment;  then  wheels 
1  and  2  are  off  the  span.  The  sum  of 


§47]  114 

their  moments  about  wheel  18  =  10  X  104 
+  20  X  96  =  2,960  and  wl  +  «?t  =  30. 
_  (14,944  —  2,960)  +  (284  —  30)  5 

~m~ 

or  R,  =  132.54.     We  proceed  similarly  for 
any  number  of  loads  off  the  span. 

47.  To  find  the  moment  about  a  point 
x  feet  to  the  right  of  p  of  all  loads  on  the 
left,  including  the  uniform  load  2  x,  let 
the  resultant  of  w^^  act  c  feet  to  left  of 
point  p  of  fig.  18,  and  call  its  moment 
about  p  =  Mp  =  ««?!_!  8  c  —  16,364;  then 
the  moment  about  the  first  point  indicated 

ry> 

is,  Mp  +  x  —  M^-IS  (c  +  *)  +  2  x  X  £ 

/.  Mp  +  x  =  16,364  +  284  x  +  x'  .  .  .  .  (19) 
From  this  formula  we  find  the  following 
values: 

Mp+  5  =  17,809,  Mp  +  25  =  24,089 
=  19,304,  Mp  +  so  =  25,784 
-  20,849,  Mp  +  35  ==  27,529 
=  22,444,  Mp  +  4o  =  29,324 


48.  Reaction  when  car-load  is  on  Bridge. 
When  the  car-load  extends  a  distance  2  to 
left  of  right  abutment  the  left  reaction  =•• 


115  [  §  48 

Mp  _t_x  -r-  I,  if  none  of  the  loads  1,  2  etc., 
are  off  the  span.  In  case  wheels  1,  2  and 
3  (say)  are  off  the  span,  their  moments 
about  right  abutment  must  be  subtracted 
from  Mp  +  x.  Thus,  for  a  120-feet  span 
and  25  =•  x  feet  of  car- load  on  the  bridge, 
the  wheels  1,  2  and  3  will  be  found  to  be 
134,  126  and  121  ft.  respectively  from  the 
right  abutment.  The  sum  of  their  moments 
about  it  consequently  —  10  X  134  -f-  20 
(126-f-121)  ==  6280;  and  this  subtracted 
from  Mp  +  25  =  24080  gives  17809,  the  mo- 
ment  of  the  loads  on  the  span  about  the 
right  abutment.  Dividing  this  by  I  =  120 
gives  the  left  reaction  R  ==  200.7. 

In  examples  of  this  kind,  mark  the  dis- 
tances along  a  straight  line,  and  it  will  tenc 
to  avoid  mistakes. 

Example  I. — For  a  span  of  120  feet,  p-is  15  feet 
to  left  of  right  abutment ;  compute  K. 

Example  II.  — For  120  feet-span,  p  is  22  feet  to 
left  of  right  abutment ;  find  E. 

Example  III.  —With  span  =  60  feet,  and  right 
abutment  2  feet  to  right  of  wheel  14,  find  left 
reaction. 


§49]  116 

49.  Moment  About  an  Apex  on  Loaded 
Chord. — An  example  will  sufficiently  illus- 
trate the  method  to  follow  in    any  case. 
Take  a  span  of  100  feet,  and  suppose  wheel 
7  at  an  apex  40  ft.   from  left  abutment; 
then  wl&  is  4  ft.  to  left  of  right  abutment. 

The  left  reaction  —  (12,041  +258  x  4) 
-=-  100—130.73  /.  moment  of  R  about 
apex  =  130.73  X  40  =  5229.2.  The 
moment  of  loads  wl__7  about  apex  or 
wheel  7  =  2,155  (from  diagram).  There- 
fore, the  moment  of  reaction  and  loads  left 
of  apex,  about  apex  as  a  centre  of  moments 
=  5229.2  —  2155  =  3074.2.  In  case  any 
loads  are  off  the  span,  R  is  found  as 
explained  in  arts.  46  and  48,  and  the 
minus  moment  by  subtracting  from  the 
tabular  value  the  sum  of  the  moments  of 
the  loads  off  the  bridge  about  the  apex 
considered. 

Example  I. — For  a  78  feet-span  find  the  moment 
at  the  centre  when  wheel  12  rests  there. 

Example  II. — Do  the  same  fora  100  feet-span. 

50.  Moment  Diagram. — In   fig.   19  is 
shown  a  moment  diagram,  constructed  by 


117 


[§50 


laying  off  the  wheel-loads  1,  2,  .  . . ,  in 
position  to  scale,  draw  ing  verticals  through 
1,  2,  . . . ,  to  intersection  with  a  horizontal 
base-line  v  t.  From  this  base,  the  mo- 
ment about  any  wheel  of  all  loads  to  its 
left  are  taken  from  the  diagram  fig.  18,  and 
laid  off  upwards  to  scale.  Straight  lines 
connect  the  points  thus  found,  forming  the 
moment  diagram  v  sf  b  r. 

I   2345   6789   10  If 

9  Q999  9090  9  Q 


Fig,  19, 


§50]  118 

Another  method  of  constructing  the  dia- 
gram is  to  compute  the  moment  of  each 
wheel  1,  2,  . . . ,  about  some  point  as  t  near 
the  right  end  of  the  sheet  and  lay  off  t  m 
=  moment  of  wl  about  t,  m  n  =  moment 
of  w^  about  t,  n  o  =  moment  of  ws  about  t, 
and  so  on.  Then  draw  v  m  to  intersection 
with  vertical  through  2.  From  this  point 
draw  a  line  to  n,  and  where  it  intersects 
a  vertical  through  3  draw  a  line  to  o,  and 
so  on.  The  moment  of  wheel-load  1,  about 

v  h 
h  (say)  —  m  t  —  =  </  A ;  of  wheel  2  about 

v  t 

2'  h 

h  =  m  n  ^/      =  g   z\   similarly   for    the 
&   t 

others.  Hence  the  sum  of  the  moments  of 
wi  __  4  about  h  —  h  /,  as  by  the  first  con- 
struction. In  the  same  way,  the  moment 
about  Jc  of  wi_io  —  k  b,  of  ^2—10  about  k 
—  I  b  (b  not  necessarily  lying  on  a  vertical 
through  a  wheel-load). 

The  moment  diagram  is  usually  con- 
structed on  profile  paper,  plate  B  with  a 
horizontal  scale  of  8  ft.  to  the  inch  and  a 
vertical  scale  about  1000  pound  feet  to  0.8 
inch.  Cross-section  paper  divided  into 


119  [§50 

millimeters  is  very  convenient,  with  a  ver- 
tical scale  about  one  half  the  above.  One 
sheet  will  then  contain  the  moments  for 
about  150  feet  length  of  train.  The  mo- 
ments for  trains  including  40  feet  of  car- 
load have  already  been  computed  in  Art. 
47,  and  may  be  laid  off  at  the  proper 
distances  to  form  the  parabolic  portion  of 
the  moment  curve  or  equilibrium  polygon. 

To  use  this  diagram  to  find  reactions 
and  moments,  lay  off,  on  a  strip  of  the  same 
paper  used  in  constructing  the  diagram,  the 
span  and  apices  where  moments  are  to  be 
found.  Then,  if  wheel  5  say,  is  to  be  placed 
at  the  apex  h,  slip  the  strip  along  until  h 
lies  on  the  vertical  through  wheel  5  and  the 
span  a  k  coincides  with  the  base. 

The  vertical  through  Jc  intersects 
the  equilibrium  polygon  at  b  (fig.  19); 
draw  a  b  or  place  a  string  from  a  to  b ;  then 
noting  the  points  d  and  f  where  a  b  and 
the  polygon  intersect  the  vertical  through 
5,  measure  d  f  to  vertical  scale,  and  its 
value  is  equal  to  the  moment  of  reaction 
and  loads  to  left  of  h  about  h.  The  proof 
is  easy;  the  sum  of  the  moments  of  all 


§50]  120 

wheel-loads  to  left  of  k  about  k  =  k  b  (to 
vertical  scale),  the  left  reaction  R  at  a  = 
Jc  b  -T-  a  k,  and  the  moment  of  R  about  h 
equals 

ah  7 

Jc  b  —=   =  h  a. 
a  K 

The  moment  about  h  of  loads  1,  2,  3,  4 
=  hf,  as  shown  above;  therefore  the  mo- 
ment of  reaction  and  loads  1,  2,  3,  4  about 
apex  h  of  truss  =  h  d  —  hf~fd,  meas- 
ured to  the  vertical  scale. 

If  the  span  was  c  k,  then  load  1  is  off 
the  truss,  and  the  moment  of  the  left  re. 

action  at  c  about  h  —  b  I  X  — j~  —  eg\ 

C  K> 

whence  the  moment  about  h  of  reaction 
and  loads  2,  3,  4  =  e  g  — f  g  —  ef. 

The  point  e  is  found  by  first  drawing  a 
vertical  through  c  to  intersection  s  with 
the  equilibrium  polygon;  then  e  is  the 
intersection  of  s  b  with  the  vertical  through 
5.  Similarly  we  proceed  if  other  loads  are 
off  the  truss.  The  lines  a  b  and  s  b  are 
called  closing  lines.  It  is  of  course  more 
accurate  to  measure  the  ordinates  k  b  or 
I  b  to  scale  and  to  compute  the  moments 


121  [ §  51 

d  h  or  e  g  by  multiplying  by  the  ratios 
ah  -4-  a  kor  c  h  -f-  c  k  ;  then  the  moments 
to-  be  subtracted  h  f  or  f  g  can  be  taken 
directly  from  the  table  (fig.  18),  only  sub- 
tracting in  the  last  case  wi  X  v  h  from  the 
tabular  moment  to  find/*  7. 

Example  I. — For  a  span  a  k,  suppose  k  b  to 
scale,  reads  5240  and  that  a  h  -r-  a  k  —  f ;  then 
d  h  =  J  (5240)  =  2328.9;  .'.  df  =  dh  -fh 
=  2328.0  -  830  (from  table)  =  1498.9  =  moment 
about  Ji  of  reaction  at  a  and  loads  1,  2,  3  and  4. 

Example  II. — Construct  an  equilibrium  polygon 
on  profile  or  cross-section  paper  to  scales  above 
for  about  150  feet  length  of  combined  engine  and 
car-load. 

POSITION    OF    LIVE    LOAD    GIVING    MAXI- 
MUM MOMENT. 

51.  The  following  investigation  applies 
either  to  a  beam  or  to  a  single  intersection 
truss  with  vertical  members  at  each  apex. 
In  figure  20  let  wn  first  rest  at  B  and  sup- 
pose that  when  all  the  loads  are  shifted  to 
the  left  a  distance  x^  the  head  of  the 
uniform  load  (p  per  foot)  is  at  the 
right  abutment  C,  so  that  no  new  load  has 


§  51  ]  122 

as  yet  come  on  the  span  and  wn  +  i  is  q 
feet  to  right  of  B. 

Let  a  =^  distance  from  wn  to  wu  +  i  .'.  a 
r=  ^  -f  ?. 

P  =    (wi    -f-   W2   +    •  •  •  ,+  *°n)     —    Sum    Of 

wheel- loads  at  and  to  left  of  B  when  wn  is 
atB. 

W  =  sum  of  all  loads  on  span  when  wn 
is  at  B. 

d  =  panel  length,  I  =.  span,  b  =  A  B, 
y  z=r  a  variable  distance  less  than  ^. 


n 


v-£ ^  I *^ 

y\/    I 

: h     \B  \w 

/?D"~'  !I     : 

Fig.20, 


Now,  beginning  with  an  initial  position 
of  the  loads  so  that  wn  rests  at  B,  on  shift- 
ing all  loads  to  the  left  a  distance  x^  -\- 
yy  the  left  reaction  R  is  increased  by 

w  *i_+_y  ,  py^ 


123  [  §  51 

and  its  moment  about  B  by 


The  minus  moment  of  P  about  B  is 
increased  by  P  (xl  -f  y)  ;  hence  the  mo- 
ment of  R  and  P  about  B  is  increased  by 
the  shifting, 


If,  in  place  of  the  uniform  load,  a  new 
wheel-load  Q  just  reaches  C  when  the 
loads  have  been  shifted  a  distance  xl3  the 
left  reaction  is  increased  on  shifting  y 
farther  to  the  left,  by  Q  y  -f-  /,  and  its 
moment  about  B  by 


so  that  the  term  in  (20)  in  the  [     ]  involv- 

ing JP  is  replaced  here  by  —  ——. 

i    i   y 

The  following  conclusions  refer  either  to 
(20)  or  to  the  formula  obtained  from  (20) 
by  the  last  substitution. 


§  51  1  124 

We  note  in  either  case  that 

1 
n  i 

—  —.  —  =   x  increases  with  y. 

- 


It  is  evident  now  from  (20)  that: 

(1)  So  long  as  the  [     ]  is  plus  for  y  =• 
o9  it  increases  with  y  and  is  a  maximum 
for  y  =  g,  so  that  wn  +  i  rests  at  B;  when 
by  the  same  method  we  ascertain  whether 
the  moment  is  increased  on  shifting  loads 
still  farther  to  the  left,  so  that  «0n-f-2  rests 
at  B,  and  so  on. 

(2)  It  may  happen  that  the  [     ]  is  minus 
before  new  loads  come  on  the  span  or  for 
y  —  o;  still  if  it  is  positive  for  some  value 
of  y,  it  is  greatest  for  y  •=  q\  the  moment 
is  increased  by  the  shifting,  and  wn  +  1  is 
moved  up  to  B. 

(3)  If,  however,  as  new  loads  are  added 
to  those  to  the  left  of  B,  P  is  increased  so 
much  that  the  [     ]  is  minus,  even  for  y  = 
q  and  xl   -f-  y  =  a,   the  moment  is  de- 
creased   by    the   shifting,  and  the   weight 
last  supposed    at  B  before  shifting  must 
rest  at  B  for  maximum  moment. 

The  case  where  some  of  the  loads  to  the 


125  [§51 

left  of  B  move  off  the  span  during  the 
shifting  will  be  considered  in  the  example 
given  farther  on. 

From  (20),  or  the  modified  form  when  Q 
comes  on  the  span,  we  have  for  the 
bracketed  term,  when  y  =:  q,  xl  -j-  y  :==  #, 

+  m-PJ] (21) 

where, 

m  ~  (Q  )  when  the  new  load  is  a 
wheel-load  and 

#1  =  r  — —  J  when  the  new  load  moving 

on  the  span  is  the   uniform  one.     In  the 

p  a 

last  case,  when  xl  =  o,  in  -  —  ,  corre- 
sponding to  the  case  where  the  uniform 
load  is  either  on  the  span,  or  its  front  is 
at  C,  when  wn  is  at  B. 

The  criterion  then  for  ascertaining 
whether  a  given  position  of  the  loads 
corresponds  to  a  maximum  moment  at  B  is 
a  simple  one: 

Assume  some  wheel-load  wu  to  rest  at  B; 
find  P  =  wl  -f-  w2  -j-  . . .  +  tvn  and  W, 


§52]  126 

the  total  load  on  the  span  when  ivn  is  at  B; 
then  if  (21)  above  is  plus,  the  moment  is 
increased  by  moving  tvn  _L  i  up  to  B;  if 
(21)  is  minus,  wn  rests  at  B.  Hence  if  we 
assume  at  the  start  such  a  position  of  the 
loads  that  (21)  is  plus,  shift  loads  to  left. 
The  instant  (21)  becomes  minus,  the  last 
load  ivn  at  B  before  shifting  must  rest  at 
B/br  maximum  moment  at  that  point. 

•  The  above  investigation  for  maximum 
moments  and  a  following  one  for  shears  is 
taken  from  a  paper  by  the  writer  on  the 
subject  in  Van  Nostrand's  Magazine  for 
March,  1885. 

52.  Graphical  Method. — In  fig.  21  the 
wheel-loads  1,  2,  3,  ...  are  laid  off  to 
scale  of  distances,  the  proper  horizontal 
distances  apart  and  vertically  to  scale  of 
loads.  This  can  be  done,  to  a  large  scale, 
on  the  moment  diagram,  fig.  19. 

Mark  off  on  a  strip  of  paper  the  span 
A  C  of  the  beam  or  truss  and  any  point,  as 
B,  where  it  is  desired  to  find  the  position 
of  loads  giving  maximum  moment.  To 
try  wheel  3  at  B,  slip  the  strip  along  the 
base-line  until  B  is  vertically  under  3. 


127  [  §  52 

Extend  the  vertical  at  C  to  meet  the  scale 
of  loads  at  E. 


I  A\( -  b ->\B 


Fig.  21. 

Neglecting  m  in  21,  we  see  that  when 

/W       P\ 

I |  is  plus,  the  loads  are  shifted  to 

left;  when  minus,  they  are  not  shifted. 
The  [  ]  changes  sign,  for  a  continuous 
loading,  when  the  average  load  on  the 
span  is  equal  to  that  on  A  B.  For  the 

TY      Q  jn 
isolated  loads,  -y-  =  -r-~  =  tan  C  A  E  and 

I          A  O 

T)  T>      O 

-7  —   ^p   =    ^an   B  A  3;    hence    as    tan 

0  J\.  -t5 

C  A  E  <  tan  B  A  3,  203  is  not  moved 
farther  to  left.  If  we  suppose  iv%  slightly 


§  52  ]  128 

P         T3  T) 

to  the  right  of  B,  -=•  =  •— -    =  tan  BAD, 
o        A  Jt> 

which  being  less  than  tan  C  A  E,  w$  is 
moved  up  to  B  for  maximum  moment.  In 
any  case  then,  B  can  be  tried  under  any 
wheel,  and  if  a  fine  string  stretched  from 
A  to  E  cuts  the  step  in  the  load-line  at 
that  point,  the  w  corresponding  causes  a 
maximum  moment  at  B. 

If  any  loads  are  to  the  left  of  A,  the  lines 
from  E,  D  and  3  are  drawn  to  a  point 
where  a  vertical  through  A  meets  the  load- 
line,  as  is  evident.* 

Unfortunately,  this  graphical  method, 
as  it  neglects  the  influence  of  new  loads 
moving  on  the  span  where  the  loads  are 
shifted  a  to  the  left,  often  requires  that 
two  or  more  contiguous  wheels  command 
at  B..  so  that  the  corresponding  moments 
have  to  be  computed  and  compared,  to  see 
which  position  corresponds  to  maximum 
moment  at  B.  For  such  cases  the  exact 

*The  graphical  rnrt^od  outlined  above  was  given 
by  Dr.  H.  T.  Eddy  in  a  paper  in  the  Transactions  Am.  Soc. 
C.  E.,  Vol  XXII.  1890,  pp.  259-358,  where  the  whole  sub- 
ject for  maximum  moments  and  shears  is  discussed. 


129  [  §  52 

analytical  method  above  (Art.  51)  gives 
but  one  position,  which  is  conclusive.  An 
exact  result  though  can  be  obtained  by  the 
graphical  method,  by  laying  off  m  (page 
125)  to  scale,  vertically  above  E,  fig.  21, 
to  fix  a  point  E',  so  that  C  E7  =  W  +  m 
of  (21).  Then  (W  +  m)  -r-  I  =  tan  C  A 
E'  and  it  follows,  as  above,  that  when  A 
E'  cuts  D  3,  wheel  3  at  B  causes  a  maxi- 
mum moment  there. 

It  will  now  be  assumed  that  the  student 
will  construct  the  load  diagram  (fig.  18)  on 
cross-section  paper  to  a  large  scale,  so  as 
to  include  about  150  feet  of  the  load, 
writing  the  loads,  moments,  and  distances 
on  it,  as  in  the  figure;  also,  that  he 
will  construct  a  diagram  of  the  truss 
next  to  be  investigated  to  the  same 
scale.  On  placing  the  first  diagram 
below  the  second,  so  that  the  assumed 
wheel  rests  at  the  apex  to  be  considered, 
the  loads  P,  W,  Q  and  the  distances  q,  a 
and  b,  are  at  once  read  off,  the  position  of 
the  load  giving  maximum  moment  for  the 
apex  quickly  found  and  the  maximum  mo- 
ment computed.  It  should  then  be  checked 


53] 


130 


by  use  of  the  equilibrium  polygon  (fig.  19), 
as  explained  in  article  50. 

53.  Computation  of  the  maximum 
stresses  in  the  chords  of  a  Pratt  truss  for 
wheel-loads. — Let  the  span  of  the  Pratt  truss 
(fig.  22)  be  120  feet,  divided  into  6  panels 
of  20  feet  each,  the  height  of  truss  26  feet- 

As  we  have  seen  above,  a  wheel-load 
always  rests  at  an  apex  for  maximum  mo- 
ment. 

B  D  F          D'         B' 


£' 


C 


6 
Fig.2Z. 

Moment  at  C.  —  Try  wheel  2  at  C. 

On  placing  the  load  diagram  along  the 
truss  diagram,  we  see  that  the  front  of  the 
car-load  is  1  foot  to  right  of  A',  P  = 
Wl  +  w2  =  30,  I  -f-  b  =  6,  W  =  284, 
a  =  distance  between  w%  and  ws  =  5  and 
q  =  4.  Then,  since  for  the  car-load  as- 
sumed p  -f-  2  —  1,  we  have  (21)  Art.  51, 


131  [§£3 

^-—  30  X  6  >  0, 
o 

hence  w$  is  moved  up  to  C. 

Next,  assuming  ws  at  C,  the  car-load  is 
on  the  bridge,  P  —  wi_3  =  50,  W  = 
284  +  4  X  2  =  292,  a  =  5  =  q,  and 


292  -+-  5  —  50  X  6  <  0. 
Therefore,  by  Art.  51  w%  rests,  or  com- 
mands, at  C  for  maximum  moment.  For 
this  position  of  the  live  load,  x  =  4  in  e^. 
19,  Art.  47,  therefore  the  left  reaction 
R  =  (16,364  +  284  X  4  +  42)  -f-  120  = 
145.967  and  its  moment  about  C  =  R  X 
20  =  29191.  The  minus  moment  of  w\  and 
«0a  about  C  =  230  by  the  table;  hence  the 
moment  of  reaction  and  loads  to  left  of  C 
about  C  =  2919J  —  230  =  2689i 

Note.  —  As  the  heavy  wheel-loads  more  fre- 
quently command  an  apex  than  the  lighter  ones, 
it  might  be  inquired  -whether  «012  or  wls  at  apex 
C  might  not  give  a  greater  moment  than  the 
equally  heavy  load  w3  ;  but  with  wls  (say)  at  C, 
the  uniform  load  is  seen  to  replace  the  second 
locomotive  and  tender  for  the  first  position  Y,  hen 


§  53  ]  132 

ws  is  at  C,  the  left  reaction  is  thus  less,  and 
therefore  the  moment  about  C  is  smaller.  The 
average  load  per  foot  of  wheels  6,  7,  8,  9,  10  is 
about  62  -T-  32,  or  nearly  the  same  as  the  uniform 
load  2  per  foot  ;  but  for  a  girder  of  over  60  feet 
span,  it  may  happen  that  the  maximum  moment 
at  some  point  of  the  span  is  greatest  for  the 
second  engine  and  tender  followed  by  the  car-load, 
since  there  is  a  space  of  11  feet  in  front  of  wll 
with  only  a  load  10  on  it,  which  would  be  replaced 
by  a  uniform  load  of  2  X  1  1  =  22  in  all.  Both 
positions  should  then  be  tried. 

Moment  About  E.  —  Try  ic5  at  E;  then 
wig  is  1  foot  to  right  of  A'.  P  =  -w  i  _  5 
=  90,  J  -5-  &  =  ti  -T-  2  =  3,  W  =  WI-IT 
=  271,  /.  (21),  Art.  51,  gives 


>  0; 

hence  w6  is  moved  tip  to  E.  It  is  not 
necessary,  in  this  case,  as  in  the  majority 
of  cases,  to  compute  Q  q  •  -r-  a  or  the  cor- 
responding expression  for  uniform  load. 

With  WQ  at  E,  3  feet  in  length  of  car- 
load is  on  the   bridge.     P   ==   103,  W  — 


133  [  §  53 

+  3  X   2   =    290,  a  —  distance  from 
to  wi  —  5  /.  (21)  of  Art.  52  gives 


290  +-5  --  103  X  3  <  0; 
"hence   tvt>   commands   at   E  for  maximum 
moment. 

The  moment  of  the  left  reaction  about  E 
equals 


5741f. 

The   minus    moment  of   w\  —  §   =    1640 
from  the   table   .'.   max.   moment  at  E  — 


It  sometimes  happens,  for  certain  spans  and 
panel-lengths,  that  a  heavy  load  of  the  second  en- 
gine commands  at  E,  as  then  the  large  minus 
moment  of  the  first  engine  is  replaced  by  the 
smaller  moment  of  the  first  tender.  Let  us  try 
^n  at  E,  in  which  case  wl_5  are  off  the  span, 

W  +  TTi  —  P  -  =  264  +  5  —  82X3>  0;  hence 
wl2  is  moved  up  to  E.  There  is  now  40  feet  of 
uniform  load  on  the  bridge.  W  -{-  m  —  PX3  — 
274  +  5  —  102  X  3  <  0  .  •.  wl2  commands  at  E. 
With  w13  at  E,  w5  is  6  feet  to  left  of  A.  The 


§  53  ]  134 

moment  of  WI_IB  and  40  feet  of  car-load  about 
A'  =  M  ^  =  29,324  (Art.  47).  From  this  we 
must  subtract  the  moment  of  wl_6  about  A'  = 
M5  +  wl_5  X  126  (see  Art.  46)  =  830  +  90  X 
126  =  12,170,  .*.  moment  of  loads  on  span  about 
A'  =  29,324  —  12,170  =  17,154.  The  moment  of 
the  left  reaction  E  about  E  is  therefore  IZ-jL5^  ^ 
2^  =  5718.  Qd 

The  minus  moment  of  wti_11  about  wls  at  E  is 
easily  found  to  be  M12  —  (M5  +  wl_6  X  46)  = 
6708  -  (830  -f-  90  X  46)  =  1738 ;  hence  the 
moment  of  K  and  w6_ll  about  E  is  5718  —  1738 
=  3980. 

This  is  less  than  4101§,  found  above,  so  that 
WQ  at  E  gives  the  greatest  moment. 

Moment  at  G. — With  w$  at  G,  the  car- 
load is  1  foot  to  right  of  A7;  a  —  8,. 
q  --  --  7,  P  --=  WI-Q  ==  142,  I  +  b  --  -.  2, 

W  -  2c4,  .*.  (21),  284  -f  ^  —   142    X   2 

o 

>  0  arid  WIQ  is  moved  up  to  G.  Note  here 
that,  if  the  new  load  moving  on  was  neg- 
lected, either  wg  or  WIQ  could  command 
at  G,  as  (21)  would  then  be  zero. 

Next  try  WIQ  at  G  and  move  loads  4  feet 
to  left  only,  whence  w\  rests  at  A. 

Including  w,,  P  ==  152,  W  —  284  4-  14 


135  [  §  53 

c=  298,  and  in  (20),  Art.  51,  a?i  :  =  0  (as 
uniform  load  is  on  the  bridge)  and  y  —  4, 
.-,  (20)  becomes  2  [298  +  4  -  -  152  X  2] 
<  0,  so  that  apparently  WIQ  must  be  moved 
back  to  G;  but  a  farther  shifting  of  loads 
to  the  left  moves  w\  off  the  span  and 
changes  this  conclusion. 

o 

Thus  with  ivi  at  A  and  not  including 
it,  P  =  142,  W  =  274  -f  2  X  11  :  =  29<>, 
x\  =  o  and  y  =  4  in  (20)  which  moves  w\i 
to  G.  /.  the  [  ]  of  (20)  gives 

[296  -j-  4  —  142  X  2]  >  0, 

so  that  the  moment  is  increased  by  shift- 
ing to  the  left  from  the  position  w\  at  A; 
but  whether  WIQ  at  G  or  wn  at  G  gives  the 
greater  moment  can  be  determined  only  by 
computing  the  moments.  We  shall,  in 
addition,  compute  the  moment  at  G  for  w\^ 
at  G,  in  which  case  w\  and  w-i  are  both  off 
the  span. 

By  the  method  explained  in  Art.  49,  we 
find  for 

Wio  at  G,  moment  at  G  =  4569, 
Wn  at  G,  moment  at  G  =  4596.5, 
Wi2  at  G,  moment  at  G  =  4569, 


§53]  136 

so  that  the  heavy  load  Wn  commands  at 
G  for  maximum  moment. 

The  graphical  method  of  Art.  50  is  here 
a  great  saving  in  time,  as  moments  like  ef 
in  fig.  19  can  be  compared  by  use  of 
dividers.  When  the  maximum  is  thus 
found  graphically,  it  can  be  computed 
afterwards  if  desired.  The  computation 
of  moments  for  w\\  at  G  is  as  follows:  w\  is 
124  feet  to  left  of  A'  (fig.  21)  and  x  =  15; 
therefore  the  moment  of  R  about  G  is 
(Arts.  47  and  49), 

(Mp  +  16  —  Wl    X      124)  —     = 

i  (20,849  —  1240)  =  9804.5 

The  minus  moment  of  W2— n  about  G 
orwn  equals  5848  (from  table  fig.  18)  minus 
(wi  X  distance  from  tvi  to  wn  =  10  X  64 
=  640)  =  5208  /.  moment  of  reaction  and 
t02-.  11  about  G  =  9804.5  —  5208  =  4596.5. 

54.  The  maximum  chord  stresses  are 
now  found  by  dividing  the  maximum 
moments  above  by  26,  the  height  of  truss 
centre  to  centre  of  chords. 


137 


[§55 


Piece 

Loading 

Moment 

Stress 

ACE 

w3  atC 

2689.333 

103.436 

BD,  EG 
BF 

w6  atE 
wll  at  G 

4101.667 
4596.500 

157.756 
176.789 

The  stresses  are  in  thousand  pounds. 


POSITION  OF  LIVE  LOAD  GIVING  MAXIMUM 
SHEAK  IN  A  BEAM. 

55.     In  fig.  20,  Art.  51,  let  w\  rest  at  B; 
the  shear  just  to  left  of  B  is  equal  to  R  = 

W  —  ,  where  s  =  distance  of  W  (the  re- 
sultant of  total  load  on  the  bridge)  from 
C  when  w\.  is  at  B.  On  shifting  to  left  a 
distance  a,  when  w%  rests  at  B,  the  shear  is 

W^  +  Q  3._Wl 

Q  q  ~  I  being  the  increase  of  R  due  to  a 
new  wheel-load  Q  coming  on  the  span, 
where  q  is  its  distance  from  C  when  w%  i& 
at  B.  The  shear  is  increased  when 


WT 


~~W1>  0; 


§  55  (a)  ]  138 

so  that  W2  rests  at  B.  If  (W  a  +  Q  (1 
-  w\  I )  is  minus,  wi  rests  at  B  for  maxi- 
mum shear.  If  iv-2  is  found  to  move  up  to 
B,  call  Wi  the  resultant  of  all  loads  on  the 
span  when  w^  is  at  B,  and  s\  its  distance 
from  C.  Then  if  loads  are  again  shifted 
to  left  a  distance  e/i,  so  that  w%  moves  up 
to  B,  the  shear  is 

TTT   si  +  #1      .    Q  q        i  ^ 

Wi  - -~ h  ~*Y  •  -  («0i  +  102  ) 

and  since  the  shear  with  w?  at  B  is  Wi   — 

—  w19  the  difference  is 

ai          Q  q 

WIT +Y-  -w* 

when  this  is  plus,  w$  rests  at  B  for  a  greater 
shear;  when  minus,  w<i  commands  there. 

For  usual  length  of  girders  and  usual 
loads,  it  is  found  that  w\  or  iv^  (but  not  w  3  ) 
commands  at  points  for  maximum  shear. 

POSITION  OF  LIVE  LOAD  GIVING  MAXIMUM 
SHEARS   IN    ANY    SINGLE  INTERSECTION 
TRUSS.* 
55  (a).     In  fig.  23  let 

*  For  a  graphical  solution  see  larger  treatises. 


130 


[§55  (a) 


P  =  (tvi  -f-  w-2  -f-  •  •  •  +  Wn  )  —  sum  of 
wheel-loads  at  and  to  left  of  C  when  wn  is 
atC. 

W  =  total  load  on  span  when  tvn  is  at  C, 
a  =  distance  from  wn  to  wn  _L  i  =  #1  +  q, 
d  =  panel-length,  /  =  span,  N  =  total 
number  of  panels,  y  =  a  variable  distance 
less  than  q. 

First  suppose  wn  at  C,  and  that  when  all 
the  loads  are  shifted  a  distance  x\  to  the 
left  a  new  wheel-load  Q  rests  at  A7  ;  then 
on  shifting  a  farther  distance  y  to  the  left, 
the  left  reaction  R  is  increased  over  the 
value  corresponding  to  wn  at  C, 

w 


.  +  Q 


and  the  load  held  at  B  by  P— *  y"  ^- 

Jp 


O    O    A 
\A 


-  I  - 
Ficj.r 


§  55  (a)  ]  140 

The  shear  in  panel  B  C  is  thus  altered 
by  the  difference, 


**x,  +  y  d\        I 

If   the    uniform    load   just    reached    A' 
when  the  loads  were  shifted  xi  to  left,  the 

term  Q  — -  would    bo    replaced     by 

^i  i  y 

*>    4/2 

Prove  this.     Observe  in  (23) 


. 

*i  +  y) 

y 

~ 


_      _ 

that    ---  r~    -   =  x\.    ,    ^  increases  with  ?/. 

#1  +  y 

y 

By  a  consideration  of  the  [  J  in  (23), 
we  see  that  the  shear  is  increased  by  the 
shifting  when  it  is  plus,  diminished  when 
it  is  minus;  or,  more  specifically: 

(1)  If  the  [     ]  is  plus  for  y  =;  o  (when 
Q  is  at  A7),  it  increases  with  y  and  is  a 
maximum  for  y    —    q   when   wn-\-i   rests 
at  C. 

(2)  The  [     ]  may  be  minus  for  y  =  o, 
but  if  it  is  plus  when  y  has  some  other 
value   (<  q)  it  attains  its  positive  maxi- 
mum for  y  —  q  or  wu  +  i  at  C. 


141  [  §  55  (a) 

(3)  If,  however,  the  [  ]  is  minus,  even 
for  y  —  #,  the  shear  is  diminished  by  any. 
shifting;  hence  wn  rests  at  C. 

We  reason  similarly  when  the  uniform 
load  comes  on;  so  that,  in  either  case, 
noting  that  when  y  —  q,  xi  +  y  =  a,  the 
expression  to  estimate  is 

[W  +  m  —  PN]  ............  (24) 


where,  m  —  (Q  -=•)  for  a  new  wheel-load 
and,  m  =  (-—    -  )    for   the    new  car-load, 

/  \  /v    Ct   / 

which  last  reduces  to  (^r~  )  when  xi  =  o. 


The  application  is  obvious.  If  w%  ,  say, 
is  placed  at  an  apex  and  (24)  is  plus,  by 
case  (1)  above,  the  loads  are  shifted  until 
ws  reaches  the  apex.  Then  with  w%  at  the 
apex,  if  (24)  is  minus  the  shear,  by  case 
(3)  above,  would  be  decreased  by  any  shift- 
ing; hence  the  shear  in  the  panel  just  left 
of  the  apex  is  a  maximum  for  ws  at  the 
apex.  Similarly  for  other  cases.  It  gen- 
erally happens  in  practice  that  wheels  1,  2 
or  3  command  at  apices.  Some  wheel,  as, 


§  56  ]  142 

we  have  seen  above,  is  always  at  an  apex 
for  maximum  shear. 

If  the  panels  are  so  short  that  w  i ,  or  both 
Wi  and  w  2  >  pass  to  the  left  of  B  during  the 
shifting,  it  is  best  to  compute  the  shears 
for  the  different  positions  and  compare 
them  directly. 

POSITION    OF    LIVE   LOAD    FOR    MAXIMUM 
SHEARS  IN  A  PRATT  TRUSS. 

56.  Consider  again  the  Pratt  truss,  fig. 
22,  Art.  53,  of  120  feet  span,  divided  into 
6  panels  of  20  feet  each  .-.  1ST  =  6  in  (24), 
Art.  55,  which  becomes 

[W  +  m  —  6  P]... (25) 

There  is  no  need  of  considering  m  when 
W  —  6  P  >  0;  but  when  W  —  6  P  <  0, 
m  may  have  such  a  value  as  to  make  W  +  m 
—  6  P  >  0,  in  which  case  it  should  be 
computed.  When  q  =  its  maximum  value 
a,  m  —  Q  q  -i-  a  cannot  exceed  20  the 
largest  value  of  Q. 

Apex  C'.— Try  wi  at  C'  .%  P  =  10,  W 
=  70  and  (25),  [70  +  m  —  60]  >  0;  hence 
tvz  is  moved  up  to  this  apex.  If  we  con- 


143  [§56 

sider  in  the  same  way  other  apices  to  the 
left,  102  is  moved  up,  as  W  in  (25)  is  larger 
than  70  and  6  P  —  60  remains  the  same; 
hence  w\  need  not  be  tried  at  the  other 
apices.  With  w^  at  C7  (25)  becomes  [90 
+  m  —  180].  As  m  <  20  always,  90  + 
m  —  180  <  0;  hence  the  loads  are  not 
shifted  farther  to  the  left,  and  w-2  commands 
atC'. 

Apex  E7. — With  wi  at  E7,  48  feet  of  load 
is  on  the  span  and  tv^  is  at  A'.  /.  Q  ==  13, 
q  =  a  —  5,  W  =  129,  P  ==  m  +w  2  =  30 
.and  from  (25),  [129  +  13  —  180]  <  0,  so 
that  w-2  commands  at  E76 

Apex  G. — With  w-2  at  G,  68  feet  of  load 
is  on  the  bridge,  .*.  wu  —  Q  is  one  foot  to 
right  of  A7 ;  hence  q  —  4,  since  a  =  dis- 
tance between  w%  and  w$  =  5.  Also  W  ~ 
172,  P  =  30  and  from  (25),  [172  +  |  20 
—180]  >  0;  hence  iv%  at  G  will  give  a 
greater  moment  than  w  2 ,  and  the  same  hap- 
pens for  all  apices  to  the  left,  since  W  is 
greater  than  for  this  case.  Here,  the  in. 
fluence  of  the  new  load  moving  on  is  seen 
to  cause  n'z  to  move  up  to  G,  whereas  if  it 
were  neglected  W2  would  command  the  apex. 


§  57  ]  144 

Apex  E. — with  Ws  at  E,  93  feet  of  live 
load  is  on  bridge,  W  =  245,  P  —  50,  .*. 
from  25,  [245  +  m  —300"]  <  0;  hence 
103  commands  the  apex  and,  a  fortiori,  it 
commands  at  G. 

Apex  C.— Wheel  3  at  c,  gives  W  =  284 
+  4x2  —  292,  P  =  50,  a  =  5  =  q,  and 
as  the  uniform  load  is  on  the  bridge,  m  = 

^—  =  a  —  5;  hence  (25), 

[292  +  5  —  300]  <  0; 
hence  w%  again  commands  the  apex.  The 
position  for  maximum  shear  at  C  is  the 
same  as  for  maximum  moment,  Art.  53,  so 
that  if  the  first  computation  has  been 
made  it  is  not  necessary  to  make  the 
second. 

It  may  be  interesting  to  note  that  if 
there  were  (G  -f-  e)  panels  of  20  feet  each 
in  place  of  6,  7^3  would  still  command  at 
the  apex  next  the  left  abutment,  since  W 
is  increased  by  40  e  only  and  PN"  by  50  e 
over  the  last  values. 

57.  Maximum  Shears  and  Stresses  in 
Pratt  Truss. — The  position  of  the  live  load 
giving  maximum  shear  in  any  panel  having 


145  [§57 

been  found  as  above,  the  shear  is  equal  to 
the  left  reaction  minus  the  part  of  the  live 
load  on  the  panel  held  at  its  left  apex, 
which  last  is  equal  to  the  moment  of  the 
load  in  the  panel  about  the  wheel  at  its 
right  end  divided  by  the  panel-length. 
This  moment  is  taken  from  the  table,  fig. 
18.  The  method  of  finding  the  reaction 
has  been  given  in  articles  46  and  48.  The 
student  should  mark  off  the  distances  re- 
quired (by  aid  of  fig.  18)  along  a  straight 
line,  or  use  the  diagrams  mentioned,  as  in 
no  other  way  can  the  results  below  be 
made  so  plain. 

Max.  Shear  in  A  B,  fig.  22.— With  w*  at 
C,  the  reaction  was  found  above  to  be 
145.967.  The  sum  of  the  moments  of  wi  and 
^V2  about  10*  (fig.  18)  is  230,  and  this  divided 
by  20  gives  the  part  of  the  load  on  A  C 
held  at  A  =  11. 5,-  /.  shear  in  A  B  — 
145.967  —  11.5  =  134.467. 

Max.  Shear  in  B  E.  w3  is  at  E;  WIG  is 
at  A7;  shear  =  12,041  —  120  —  11.5  = 
88.843. 

Max.  /Shear  in  G  D  and  D  E.  w$  at  G; 
^/5l2  is  4  feet  to  left  of  A'. 


§57]  146 

R  =  (6708  +  192  X  4)  -f-  120  =  62.3, 
/.  shear  =  62.3  —  11.5  =  50.8. 

Max.  Shear  in  E'  F  and  F  G.  w*  is  at 
E7  and  WQ  at  A7, 

Shear  =   (3496  -4-  120)  —  (80  ~  20)   = 
25.133. 

Shear  in  E1  B'  with  w2  at  C7.— We  find 
w5  to  be  5  feet  to  left  of  A7,  /.  R  =  (830  + 
90  x  5)  -f-  120  =  10,667.  As  the  part  of 
wi  held  at  E7  =  80  -f-  20  =  4,  the  shear 
is  6.667. 

This  shear  alone  would  cause  compres- 
sion in  E7  B7.  Similarly,  with  w$  at  C  and 
load  extending  to  left  abutment,  the  com- 
pression in  B  E  =  6.667  X  see  i  =  8.414, 
as  noted  below;  and  this,  combined  with 
the  dead-load  tension  in  B  E,  gives  its 
minimum  stress. 

We  have  here,  tan  i  =  20  ~  26  =  .76923, 
/.  i  =  37°  347  and  sec  i  =  1.262. 

On  multiplying  shears  on  end  post,  ties 
and  counters  by  1.262,  their  stresses  are 
found  and  entered  below. 

The  dead-panel  load  for  this  truss  (see 
Art.  5,  eq.  5)  is  13.5,  one- third  of  which  is 
assumed  on  the  upper  chord.  The  stresses 


147 


[§57 


(tension  —  -f-,  compression  — )are  found  as 
in  Art.  38,  and  then  combined  with  the  live- 
load  stresses. 

MAXIMUM  AND  MINIMUM  STRESSES  IN  END 
POST  AND  DIAGONALS. 


Load 

AB 

BE 

DG 

E'F 

Live 

—  169.907 

-f  112.120 

+  64.110 

+  31.718 

—      8.414 

Dead 

—    42.854 

-f    25.555 

-}-     8.520 

—    8.520 

Max. 

—  212.821 

-f  137.675 

+  72.630 

-f  23.198 

Min. 

—    42.854 

+    17.141 

0 

0 

MAXIMUM  AND  MINIMUM  STRESSES  IN 
VERTICALS. 


Load 

D  E 

F  G 

BC 

Live 

—  50.800 

—  25.133 

+  65.550 

Dead 

—  11.250 

,+    2.250 

+    9.000 

Max. 

—  62.050 

—  22.883 

+  74.550 

Min. 

—    4.500 

—    4,500 

-4-    9.000 

As  the  max.  and  min.  stresses  in  B  E 
are  both  tension,  no  counter  is  required  in 
panel  C  E.  The  minimum  stress  in  D  G 


§57]  148 

nz  o  when  E  F  is  in  action  and  stress  ift 
E7  F  =  o  when  G  D7  is  in  action. 

The  dead-load  stress,  2.25  tension,  in 
F  G  occurs  when  F  E7  is  in  action,  and  is 
the  shear  in  F  G  due  to  dead  load  alone. 
It  is  the  part  of  the  dead  load  9  at  G  that 
must  go  to  the  right  abutment  to  make 
both  reactions,  at  A  and  A7,  equal ;  and  this 
part  must  travel  up  F  G,  since  G  D7  is  sup- 
posed out  of  action.  The  maximum  stress 
in  F  G  can  likewise  be  found  as  in  Art.  38. 
Thus  shear  in  F  E7  due  to  live  and  dead 
loads  =  25.133  —  6.750  ==  18.383.  The 
stress  in  F  G  =  18.383  +  4.5  (dead  load 
at  F)  ==  22.883,  as  given  above. 

The  minimum  stresses  in  D  E  and  F  G, 
given  above,  are  those  due  to  dead  loads 
at  top  of  posts  when  the  diagonals  con- 
necting with  their  tops  are  out  of  action. 

The  (minimum)  dead-load  stress  in  the 
hip  vertical  B  C  is  simply  the  dead  load 
held  at  C;  the  maximum  stress  is  deter- 
mined in  the  next  article. 

The  minimum  stresses  in  the  chords  are 
those  caused  by  dead  load. 


149 


[§58 


MAXIMUM  AND  MINIMUM  STRESSES  IN 
CHORDS.     (SEE  ART.  54). 


Load 

ACE 

BD,  EG 

DF 

Live 
Dead 

103.436 
25.960 

157.756 
41.536 

176.789 
46.728 

Max. 

129.396 

199.292 

223.517 

58.  Maximum  Live-load  Stress  in  Hip 
Vertical. —In  fig.  24,  two  end  panels  are 
shown.  As  in  skew  bridges  the  panel- 
lengths  are  often  unequal.  Let  12  =  di,  23 
=  f/2.  Also,  suppose  wn  to  rest  at  2  and 
P  =  resultant  of  w±  _  n  to  act  a  distance  c 
to  left  of  2  and  W  the  resultant  of  all  the 
wheel-loads  on  1  2  3  to  act  a  distance  s  to 
left  of  3.  Then  calling  RI  the  reaction  of 
the  floor  beam  at  1  or  part  of  the  load  on 
12  supported  at  joint  1  and  Ra  the  part  of 
the  load  on  123  held  up  by  the  hip  vertical 
2  B,  we  have,  taking  moments  about  joints 
2  and  3  of  the  reactions  and  loads  on  the 
left,  Ri  di  =  Pc;Ri  (di  +  d2 )  +  R2  d*  == 
W  s.  Substituting  in  the  last  equation  the 
value  of  Ri  obtained  from  the  first, 


§58] 


150 


R2  = 


As  before,  let  a  =  distance  from  «f'n  to 
«0n  + 1  ?  and  suppose  loads  shifted  to  left  a 
distance  x  <  a  when  a  new  wheel-load  Q 
is  supposed  to  move  on  the  panel  2  3  a  dis- 
tance y,  then  s  and  c  are  each  increased  by 
x  and  £/ie  change  in  Rj  is 


L  * 

If  this  is  plus  for  some  value  of  a?,  the 
increase  of  R2  is  greatest  for  x  =  a  when 
y  =  q  (say),  so  that  wn  +  i  rests  at  2. 
Again,  determining  P,  W,  Q  for  wn  +  i  at  2, 
if  now  (26)  is  minus  even  for  x  =  a,  y  =  g, 


151  [  §  58 

R2  is  diminished  by  any  farther  shifting, 

so  that  wn  +  1  rests  at  £  for  maximum  R  2  . 

Making  x  —  a,  y  =  q  in  (26),  we  have, 


When  tlie  [  J  is  plus,  we  shift  to  left, 
and  continue  thus  until  it  changes  to  minus. 
The  instant  this  occurs,  the  last  load  sup- 
posed at  2  remains  tiere  for  maximum  RQ. 

By  reference  to  Art.  51,  this  is  seen  to 
be  the  criterion  for  finding  what  wheel- 
load  should  command  at  2  for  a  maximum 
moment  there  when  the  span  of  the  bridge 
is  13  =  di  +  d2  . 

We  have  supposed  above  that  no  loads 
on  1  2  pass  to  the  left  of  1  ;  if  any  do,  it  is 
best  to  compute  R2  directly  for  two  or 
more  positions  of  the  live  load,  and  thus 
determine  maximum  R2  by  trial. 

The  value  of  R2  above  can  be  put  in  a 
form  very  convenient  for  computation. 
Call  M2  and  MS  the  sum  of  the  moments 
of  the  wheel-  loads  to  left  of  2  and  3  about 
2  and  3  respectively;  then  PC  =  Al2> 
Ws=  M3, 


§58]  152 


A 

To  apply  the  above  theory  to  finding  the 
maximum  live-load  stress  ever  sustained 
by  the  hip  vertical  in  the  Pratt  truss  hav- 
ing panels  of  20  feet  each,  the  [  ]  in  (27) 
becomes,  since  d\  =  d%9 

[W  +  Q  |   -  -  2  P] (29) 

Suppose  ws  at  2,  then  W  —  WI  —  Q  = 
103,  P  =  50,  /.  W  —  2  P  >  0  and  w±  at  2 
gives  a  greater  R2.  For  this  position  of 
the  loads,  W  =  wi_7  =  116,  P  =  70  and 
no  new  load  moves  on  when  w±  is  shifted 
5  feet  to  left.  [W  —  2  P]  <  0,  /.  w±  rests 
at  joint  2  for  maximum  R2. 


..    ..(30) 


By  use  of  table,  fig.  18,  M3  =  2155  + 
116  x  1  ==  2271,  also  2  M2  =  2  X  480 
=  960  and  d  =  20,  .*.  R2  ~  65.55,  as  given 
in  the  preceding  article. 

When  new  loads  moving  on  are  neg- 
lected, we  note  that  (29)  shows  that,  for 


153  [  §  59 

equal  panel-lengths  the  loads  on  the  two 
panels,  if  continuous,  should  be  equal. 

If  there  are  two  lines  of  stringers,  as  is 
usual,  the  maximum  moment  and  shear  for 
a  floor  beam  correspond  to  two  loads  of 
€5,550  pounds,  each  applied  at  the  points 
where  the  stringers  are  riveted  to  the  floor 
beam. 

59.  Maximum  Moment  in  Stringer. 
Let  P  —  resultant  of  loads  at  B  and  to 
its  left,  fig.  25,  and  c  its  distance  from  B, 
W  —  resultant  of  all  wheel-loads  on 
stringer  A  D  and  g  its  distance  from 
wheel  at  B.  Then  as  the  distance  A  B 
=  x  varies,  the  loads  shifting  at  the  same 
time,  so  that  the  wheel-load  first  supposed 
at  B  remains  always  at  B,  the  moment 
at  B  varies  and  must  have  a  maximum  for 
some  value  of  x. 


InH 

t      r 


O     *'  '  O  O  O 


1           III 

fl 

&7    . 

f 

* 

Fi3.25. 

1 

§59]  154 

This  moment  at  B  is 

w 

M=ry  (l  —  x—  g)x—  PC  ......  (31) 

For  a  maximum,  its  derivative  with 
respect  to  x  must  be  zero. 

.'.  (I  —  2  x  —  g)  —  o,  or,  ^  +  —-  =  —- 

Therefore  the  maximum  moment  under 
the  particular  wheel  at  B  occurs  when  the 
load  at  B  is  as  far  from  the  centre  C  of  the 
beam  on  one  side  as  the  centre  of  gravity 
of  the  loads  on  the  beam  is  on  the  other 
side  of  the  centre. 

Article  52  shows  that  wheels  3  or  4 
command  at  C  for  maximum  moment  there, 
for  I  —  20,  when  only  the  heaviest  wheel 
loads  2,  3,  4  and  5  are  on  the  stringer, 
hence,  wheel  3  will  be  tried  at  B  for  the 
greatest  of  all  possible  moments  at  any 
point  of  the  beam. 

To  find  g,  take  moments  of  w<z,  w^, 
w±<,  105  and  W,  about  103. 

W  g  =  W4,  X  o  +  w5  X  10  —  w<>  X  5 


155  [§59 

Hence,  w%  is  placed  1.25  feet  to  left  of 
C,  in  which  case  W  would  be  1.25  feet  to 
right  of  C. 

The  moment  about  B  is  then  easily  found 
to  be  206.25.  The  same  result  follows  if  the 
loads  are  reversed,  and  w±  is  taken  at  B ; 
hence,  this  is  the  greatest  moment  ever  ex- 
perienced at  any  point  of  the  stringer  as 
the  rolling  load  moves  over  it  in  either 
direction.  The  maximum  moment  is  most 
readily  found  by  substituting  the  value  of 
x  =  %  (I  —  g]  corresponding  to  it,  in  (31) 
above,'  giving 

W 

max.  M  =  ^  (I  —  g)*  —  PC (32); 

or  for  this  case 

CA 

max.  M  ==  ^^  (17.5)2      -  20   X   5    = 

206.25  as  before. 

When  WB  is  at  C,  the  moment  there  is 
200,  or  less  than  the  above. 

The  maximum  shear  in  the  stringer  oc- 
curs when  W2  is  at  A,  whence  105  is  5  feet 
from  D.  The  shear  then  is  equal  to  the 
left  reaction,  and  is 


§60]  156 

20 

2-  (5  +  10  +  15  +  20)  =  50. 

Example. — Find  the  maximum  moment  in  a 
floor  joist,  20  feet  long,  sustaining  two  road  roller 
loads  of  3200  and  2880  pounds  respectively,  11 
feet  apart. 

60.  Position  of  Load  for  Maximum 
Moment  in  a  Warren  Girder. — In  Art.  51 
the  method  was  given  for  finding  the  po- 
sition of  live  load  corresponding  to  maxi- 
mum moment  at  any  point  of  a  beam,  the 
Pratt  or  Howe  truss,  or  the  Warren  with 
sub- verticals,  fig.  5.  The  same  method 
evidently  applies  to  the  Warren  proper, 
when  the  centre  of  moments  is  taken  on 
the  chord  that  bears  the  roadway,  but  not 
when  the  centre  is  taken  on  the  other  chord. 
The  strict  investigation  is  then  briefly  as 
follows: 

In  fig.  26,  let  the  roadway  be  on  the 
lower  chord,  and  suppose  it  is  desired  to 
find  the  loading  that  will  cause  a  maxi- 
mum stress  in  D  E.  B  is  the  centre  of 
moments.  Call  the  horizontal  distances 
from  A  and  D  to  B,  b  and  c  respectively; 


157 


§60] 


also  let  Wi  —  resultant  of  loads  on  A  D,  in- 
cluding any  load  at  D,  W2  =  resultant  of 
loads  on  D  E,  including  any  load  at  E  and 
Wa  =  resultant  of  loads  on  rest  of  span,  E 
A'.  Let  W  =  Wi  +  W2  +  W3  and  sup- 
pose  no  load  to  move  off,  but  a  new  load 
Q  to  move  on  the  span,  a  distance  ?/,  when 
all  the  loads  are  shifted  to  the  left  a  dis- 
tance x,  in  which  case  R  is  increased  by 


and  its  moment  about  B  by 


>  J*     jsr'- 

Fig.26. 

•T+«.f> 


The  minus  moment  of  Wi  about  B  is  in- 
creased by  Wi  x. 

The  amount  of  W^,  transferred  by  the 
stringers  to  D  is  W2  t  ~  d\  this  is  increased 


§60]  158 

by  the  shifting  (if  no  loads  pass  D  or  E), 
W2  x  -f-  d  and  its  moment  about  B  by 
W2  c  x  -r-  d.  The  total  moment  about  B 
is  thus  altered  an  amount  equal  to 


If  this  expression  is  plus  for  x  =  e, 
y  •=.  q7  when  a  load  reaches  D  or  E,  the 
moment  is  increased  by  the  shifting;  if 
minus,  it  is  decreased.  This  expression, 
for  x  —  e,  y  =  q,  can  be  put  in  the  form, 


hence  if  the  [  ]  here  is  plus,  we  shift 
until  a  load  reaches  D  or  E,  and  continue 
thus  until  the  [  ]  changes  to  minus. 
When  this  first  occurs,  the  last  load  sup- 
posed at  D  or  E  remains  there  for  maxi- 
mum moment  at  B. 

When  a  uniform  load  p  per  "foot  moves 
on  the  span  a  distance  q  when  the  loads 
are  shifted  to  left  x  =  e,  the  term  Q  q  -f-  e 

above  is   replaced  by  (^-  —  —  ),  which  be- 

\  /w      €>   I 

comes  (~  e\  when  q  =  e. 


159  [§61 

On  comparing  the  above  bracketed  term 

with  (21)  Art.  51,  we  note  that  P-  of  (21)  is 

(c  \  I 
Wi  -f  Wo  —j  j  T>  which  be- 
comes identical  with  the  former,  for  c  =  o, 
as  should  be  the  case. 

For  the  Warren  girder,  c  -f-  d  =  0.5. 

The  above  investigation  applies  also  when 
the  upper  chord  is  inclined  as  in  the  Bow- 
string, Pegram  and  other  trusses,  as  the 
shape  of  the  unloaded  chord  does  not  en- 
ter into  the  investigation. 

Example  I. — Find  the  position  of  the  loads  (fig. 
18)  that  causes  a  maximum  moment  in  the  third 
loaded  chord-panel  of  a  through  Warren  girder 
(fig.  1)  of  108  feet  span  divided  into  6  panels. 

Example  II. — For  the  same  system  of  loads, 
find  the  maximum  live-load  stresses  in  a  Pratt 
deck  bridge  having  5  panels,  each  22  feet  long, 
the  height  of  truss  being  24  feet,  giving  the  posi- 
tion of  the  loads  for  each  case. 

61.  Equivalent  Loads. — The  method  of 
finding  maximum  stresses  for  the  actual 
wheel-loads,  proving  very  laborious  in  prac- 
tice, attempts  have  been  made  to  substitute 


§61]  160 

a  so-called  "equivalent"  uniform  load, 
either  alone  or  in  connection  with  one  or  two 
locomotive  excesses.  Such  loads  are  found 
as  will  give  the  same  stresses  in  certain  mem- 
bers as  the  actual  wheel-loads  and  these 
loads  are  then  assumed  for  all  the  other  mem- 
bers of  the  bridge.  If  one  locomotive  ex- 
cess is  used,  it  is  placed  at  the  head  of  the 
uniform  load  for  maximum  shears;  if  two 
excesses  are  used,  the  second  is  placed 
about  50  feet  or  more  back  of  the  first. 
For  maximum  chord  stresses,  the  uniform 
load  covers  the  span,  and  one  locomotive 
excess  is  placed,  in  addition,  at  the  apex 
about  which  moments  are  taken. 

As  to  the  agreement  by  the  two  methods, 
the  reader  may  consult  the  elaborate 
papers  by  J.  A.  L.  Waddell  in  Transac- 
tions of  the  American  Society  of  Civil  En- 
gineers ;  Vol.  XXVI.,  p.  71,  and  Vol.  XXXI., 
p.  213,  also  Johnson's  Framed  Structures, 
chap.  VI.,  where  the  possibilities  and  draw- 
backs of  the  approximate  methods  are  fully 
discussed. 


1G1  [§62 

WIND  PRESSURE. 

62.  The  wind  is  assumed  to  blow  at 
right  angles  to  the  bridge,  and  to  exert  a 
maximum  pressure  of  30  pounds  per  square 
foot  against  the  sides  of  both  trusses,  the 
entire  flooring  system  (including  track  and 
hand-railing  if  any)  and  a  train  of  cars 
also  in  case  of  a  railroad  bridge.  A  force 
of  30  pounds  per  square  foot,  and  even  less, 
will  overturn  an  empty  freight  car,  and 
thus  wreck  a  bridge  loaded  with  cars, 
which  accounts  for  the  limit  imposed.  For 
highway  bridges  this  limit  is  sometimes 
exceeded,  though  the  pressure  above,  cor- 
responding to  a  hurricane,  is  rarely  expe- 
rienced in  the  life-time  of  any  metal  bridge. 

It  is  customary  to  conceive  a  plane  mid- 
way between  the  upper  and  lower  chords 
of  a  bridge  and  to  regard  the  upper  lateral 
system  as  carrying  all  the  wind  pressure  on 
the  parts  above  the  plane  and  the  lower 
lateral  system  all  the  wind  pressure  on  the 
parts  below  this  plane. 

The  total  area  exposed  to  the  pressure 
above  is  equal  to  the  sum  of  the  vertical 


§  63  ]  162 

projections  on  a  plane  parallel  to  a  truss  or 
perpendicular  to  the  direction  of  the  wind 
of  all  parts  of  the  bridge,  including  both 
trusses. 

To  avoid  this  computation,  a  common 
specification  now  is  to  allow  150  pounds 
pressure  for  each  foot  of  span  to  both  the 
upper  and  lower  horizontal  lateral  trusses, 
to  be  treated  as  dead  load  ;  also  on  high- 
way bridges,  150  pounds  per  foot  of  span 
on  the  loaded  chord  to  be  treated  as  mov- 
ing load  and  for  railroad  bridges  300 
pounds,  similarly  treated.  This  last  corre- 
sponds to  30  pounds  pressure  on  a  car  body 
10  feet  in  height.  The  dead  load  above 
is  sometimes  increased  0.4  pounds  per  lin- 
eal foot  for  each  additional  foot  in  length 
of  span  over  200  feet,  150  pounds  being 
allowed  up  to  200  feet  span. 

Let  us  use  this  approximate  rule  for  the 
Pratt  truss,  fig.  27  (a)  of  6  panels  of  20  feet 
each,  the  depth  of  truss  being  20  feet. 
Figs,  b  and  c  are  the  horizontal  projections 
of  the  upper  and  lower  horizontal  bracing 
between  similar  chords  of  the  two  trusses. 

63.     Lateral  Trusses. — For   the   lateral 


163 


[§63 


truss  between  the  two  upper  chords  we 
have  by  the  rule  above,  the  panel-loads  20 
X 150— 3000  pounds  or  1500  pounds  at 
each  panel-point  b  to  b'  and  B  to  B',  the  load 
being  supposed  divided  between  the  two 
chords,  as  the  wind  pressure  on  both 


i     i 


(a) 


(c) 


r  r  r 

F  £>'          B' 


C            E           6           E'          C' 
a         c  e          9 & c' 


A  C  EG  E'          C         A' 

Fig.27. 

trusses   is  included.     The   pressures  at  B, 
B7,  b,  b'  are  each  about  1500  as  at  the  other 


§63]  164 

apices,  since  the  pressure  on  half  the  batter 
braces  (end  posts)  is  carried  there.  If  this 
wind  load  is  supposed  to  act  from  /'  tow- 
ards F,  fig.  (#),  the  full  diagonals  act  ;  if 
in  the  other  direction,  the  dotted  diagonals. 
For  the  first  case,  the  loads  are  carried  by 
the  lateral  trusses  to  B,  &,  B',  b',  and  there 
transferred  by  aid  of  the  portal  bracing  and 
end  posts  to  the  abutments.  Leaving  off  the 
loads  at  the  ends,  the  total  wind  load  trans- 
ferred by  the  bracing  to  b  or  b1  is  4500 
pounds,  or  half  the  load  acting  at  interior 
points.  Including  now  that  at  the  ends, 
we  see  that  the  portal  bracing  has  to  be 
designed  to  sustain  loads  at  b  and  b' 
of  6000  pounds,  and  at  B  and  B'  of  1500 
pounds  each.  With  the  wind  blowing  in 
the  opposite  direction,  these  loads  are  in- 
terchanged. 

The  shears  for  wind  blowing  from  / 
towards  F  are:  D  b  =  4500,  D  d=  3000,  F 
d  =  1500,  /  F^1500,  from  which  the 
stresses  in  web  members  and  chords  are 
found  for  a  distance  between  trusses  of  16 
feet,  centre  to  centre,  as  follows  : 


165  [ §  63 

D  1}  —  +  5625,  D  d  =  —  3000, 
Fd  =  :  +  2400,  fF  =    -1500, 
B  d  =  o,  D  F  =  +  5625, 
ld  =     -5625,  J/=     -7500. 
The  chord  on  the  windward  side  is  thus 
in  compression,  the  leeward  chord  in  ten- 
sion. 

For  the  lower  lateral  truss  (c)  we  have,  as 
before,  1500  pounds  fixed  load  at  each  inter- 
mediate apex,  and  for  a  railroad  bridge 
3000  pounds  moving  apex  load.  Both  are 
supposed  to  be  applied  at  the  joints  of  both 
chords.  Wind  is  supposed  to  blow  in  di- 
rection g  G. 

The  dead-load  stresses  are  : 
a  C  =  +  12,000,  cE;=:  +  7200,6  G  = 
+  2400. 

c  C  -     -  6000,  e  E  --     -  3000,  g  G  = 

-  1500. 

A  c  •=  0,  C  E  =  -{-  9375,  E  G  =  + 
15,000. 

a  c  =    -  9375,  c  e  -      -  15  000,  e  g  = 

-  16,875. 

For  full  live-load  only  the  chord  stresses 
are  double  the  above.  For  maximum  shears 
due  to  moving  load,  (14)  of  Art.  32  gives 


§G4] 


1GG 


the  shears  on  the  diagonals,  S  =  (6  —  n) 
(7  —  n)  X  500;  therefrom  the  maximum 
stresses  due  to  moving-load  in  ties  and 
posts  are  as  follows: 

o  €  =  +  24,000,  c  E  =  +  16,000,  e  G 
==  +  9600. 

c  C  =  -  13,000,  e  E  =  —  9000,  g  G  = 
—  6000. 

For  live  and  dead-loads  combined,  g  G 
bears  a  stress  =  -  6000,  as  then  counter 
g  E7  is  in  action,  and  dead-load  shear  in  g 
G  =  0.  The  stresses  in  thousand  pounds 
are  then  as  follows: 


aC 

c  E 

eG 

cC 

0E 

9& 

Wind  on  truss. 

+  12.0 

+    7.2 

+  2.4 

—    6.0 

—  3.0 

0.0 

"      "  train. 

+  24.0 

+  16.0 

+  9.G 

-13.0 

—  9.0 

—  6.0 

The  chord  stresses  above  are  entered  in 
the  table,  Art.  66,  opposite  "wind  on 
truss"  and  "wind  on  train"  for  both 
directions  of  the  wind. 

64.  Overturning  Action  of  Wind  on 
Truss. — In  fig.  28  is  shown  a  cross-section 
of  the  bridge  and  of  a  car  on  the  rails. 


167 


[§64 


The  total  wind  force  on  the  windward 
upper  chord  is  5  X  1500  =.  7500  pounds, 
and  the  same  force  acts  on  the  leeward 
chord  as  marked  on  the  figure.  These 
forces  tend  to  overturn  the  bridge,  the 
axes  of  rotation  being  the  roller  or  other 
supports  on  the  leeward  side.  This  axis  is 
a  little  below  the  lower  chord;  but,  as  the 
wind  really  acts  a  little  below  the  top 
chord,  call  the  arm  26  feet,  the  depth  of 
truss.  If  we  call  x  the  increased  pressure 


Fig.28. 

at  the  two    leeward   supports,    this   same 
amount   will    represent   the    decrease    of 


§65]  168 

pressure  at  the  windward  supports,  for 
taking  moments  about  either  support  we 
have  15,000  X  26  =  16  x 

.'.  —  =  12,187 

/v 

gives  the  vertical  pressure  at  one  leeward 
support  or  the  decrease  of  pressure  at  one 
windward  support.  This  vertical  force 
then  acts  as  a  reaction  on  the  end  post 
A  B,  fig.  27  (a),  and  causes  a  stress  in  it 
=  12,187  X  1.26  =  15,356,  where  1.26  is 
the  secant  of  the  angle  it  makes  with  the 
vertical.  The  tangent  of  the  same  angle 
=  0.769;  hence  the  stress  in  the  chord  A 
A'  equals  12,187  X  .769  =  9371. 

On  the  leeward  truss,  this  gives  com- 
pression in  end  post,  and  tension  in  the 
lower  chord;  in  the  windward  truss 
these  stresses  are  reversed.  It  is  seen 
that  this  increase  or  decrease  of  pressure 
in  chords  is  uniform  from  end  to  end. 

65.  Overturning  Action  of  Wind  on 
Train.— Let  us  suppose  a  car  body  10  feet 
high,  on  which  the  wind  exerts  a  pressure 
of  30  pounds  per  square  foot.  For  a  panel- 
length,  or  20  feet  length  of  car,  the  force 


169  [§66 

of  the  wind  is  P  =  10  X  20  X  30  — 
6000.  The  centre  of  pressure  on  a  freight 
ca~  is  about  7  feet  above  the  rails;  call  it 
11  feet  4  inches  above  the  centre  of  the 
lower  chord.  If  x1  denotes  the  increase  of 
pressure  per  panel  on  the  leeward  truss , 
and  a  like  decrease  of  load  on  the  wind- 
ward truss,  we  have 

16  x1  =  6COO  X  Hi 
/.  x1  =  4250. 

Regarding  this  load  per  panel  as  a  mov- 
ing-load, we  find  the  maximum  stresses 
due  to  it  in  the  leeward  vertical  truss  (Art. 
32) ;  A  B  -  -  13,380,  B  E  =  +  8920, 
D  E  =  —  4250,  D  G  =  +  5355,  F  G  = 
-  2125,  F  E7  =  +  2677,  E7  B'  =  -  892. 

For  the  panel-load  x'  =  4250,  at  each 
lower  apex  we  find  the  chord  stresses  in 
the  vertical  truss. 

A  C  E  =  +  8175,  E  G  =  +  13,080,  B  D 
-  13,080,  D  F  =      -  14,715. 

For  the  wind  blowing  in  the  opposite 
direction,  the  signs  of  all  these  stresses  are 
changed. 

6G.  These  stresses  are  entered  in  the 
following  tables  under  the  head  "wind 


§  60  ] 


170 


overturning."  We  may  conceive  the  bridge 
to  lie  north  and  south,  A  being  the  north 
end,  so  that  when  the  wind  blows  from 
the  east,  ABB7  A7  is  the  leeward  truss, 
and  when  from  the  west  the  windward 
truss.  The  live  and  dead- load  stresses 
found  in  Art.  57  for  the  same  truss  are 
given  first  in  the  tables,  and  then  com- 
bined with  the  stresses  due  to  wind,  in 
order  to  get  the  total  maximum  and  mini- 
mum corresponding  to  the  most  adverse 
conditions  simultaneously  occurring. 


Main  Ties 

Coun- 
ter 

Verticals 

BE         D  G 

FW 

BC 

D  E 

FG 

1 

Dead  load 

+  25.6 

+    8.5 

—    8.5 

+    9.0 

—  11.3 

+    2.3 

Live  load 

+  112.1 

+  64.1 

+  31.7 

+  65.6 

—  50.8 

—  25.1 

Wind  over- 

turning : 

On  train 

from  E 

+     8.9 

+    5.4 

+    2.7 

+    4.3 

—   4.3 

-    2.1 

On  train 

from  W 

—     8.9 

-    5.4 

—    2.7 

-    4.3 

+    4.3 

+    2.1 

Max.  stress 

+146.6 

+  78.0 

+  25.9 

+  78.9 

—  66.4 

—  24.9 

Min.  stress 

+  16.3 

0 

0 

+    9.0 

—   4.5 

—  4.5 

To  save  space,  the  stresses  are  taken  to 
the  nearest  100  pounds,  and  are  then  all 


171 


[§66 


expressed  in  thousand-pound  units,  +  for 
tension,  —  for  compression. 

The  minimum  stress  in  B  E  is  the  same 
as  that  in  B'  E7  or  (Art.  57),  —  8.414  from 
live  load,  +  25.555  from  dead  load  and 
(Art.  65)  —  .892  from  wind  overturning 
(one  panel  C'  loaded)  —  16.3. 


Batter 
Brace 

Upper  Chord 

Lower  Chord 

A  B 

BD 

DF 

AC 

CE 

EG 

Dead  load  
Live  load  
Wind 
on  truss  fromE 

on  train  from  E 

Do.  Overtum'g: 
on  truss  fromE 

on  train  from  E 
"     "      "    W 

Maxim,  stress 
Min.  stress 

—  42.9 
-170.0 

—  41.5 
—157.8 

0 
-    5.6 

—  46.7 
—176.8 

+    5.6 
—    7.5 

+  26.0 
+103.4 

0 
—    9,4 
0 
—  18.8 

+    9.4 
—    9.4 

+    8.2 
—    8.2 

+147.0 
+    7.2 

+  26.0 
+103.4 

+    9.4 
—  15.0 
+  18.8 
—  30.0 

+    9.4 
—    9.4 

+    8.2 
—    8.2 

+175.2 
+     1.2 

+  41.5 
+157.8 

+  15.0 
—  16.9 
+  30.0 
—  33.8 

+    9.4 
—    9.4 
+  13.1 
—  13.1 

+266.8 
+  15.2 

-15.4 
+  15.4 
—  13.4 
+  13.4 

—241.7 
—  27.5 



-    13.1 
-   13.1 

-212.4 
-   41.5 

-    14.7 
+    14.7 

-232.6 
-    41.1 

In  computing  maximum  and  minimum 
stresses  from  the  tabular  values,  be  careful 


§66]  172 

to  combine  only  those  stresses  which  must 
occur  at  the  same  time 

Thus  the  maximum  stress  in  D  F  is  due 
to  "dead  and  live  load"  (—46.7  —  176.8), 
"wind  overturning"  on  truss  and  train 
from  east  (—  14.7)  and  "wind  on  truss" 
from  east  (-{-  5.6),  giving  in  all  a  stress  of 
(-232.6). 

In  finding  the  minimum  for  D  F   no 

train    is   supposed    on    the   bridge,    and 

the   wind    from  the    east  is  seen  to  give 

the     least     stress    =    -  -   46.7  +  5.6   = 

-41.1. 

The  stresses  in  the  upper  lateral  bracing 
(Art.  63)  are  seen  to  be  very  small.  In 
cases  of  this  kind,  the  members  are  de- 
signed arbitrarily,  no  section  being  al- 
lowed less  than  three-fourths  of  a  square 
inch.  Stiff  lateral  bracing  is  of  material 
aid  in  preventing  vibration.  In  the  lower 
system  the  floor  beams  usually  act  as  the 
struts,  the  diagonals  being  suitably  fas- 
tened to  them. 

In  the  last  table  above  it  will  be  noticed 
that  the  wind  from  the  west,  on  the  empty 
truss,  comes  near  reversing  the  stress  in 


173  [§66 

the  second  panel  C  E  of  the  lower  chord. 
In  cases  where  this  happens  the  panels 
where  reversion  can  occur  should  be  de- 
signed to  act  as  struts  as  well  as  ties, 
though  sometimes  the  stringers  are  relied 
on  to  take  the  exceptional  compression. 

Some  engineers  take  no  notice  of  the 
stresses  due  to  wind  in  the  vertical  trusses 
when  they  aro  less  than  those  due  to  live 
and  dead-loads,  unless  they  cause  a  rever- 
sal of  stress  in  the  lower  chord,  the  reason 
being  that  such  stresses  are  extremely  un- 
likely to  occur  in  the  lifetime  of  the  bridge, 
and  that  when  they  do  occur  they  never 
strain  the  metal  up  to  the  elastic  limit  on 
account  of  the  low  unit  stresses  taken 
for  ordinary  dead  and  live-load  stresses. 
Therefore,  it  is  claimed  that  such  excep- 
tional stresses  can  be  safely  allowed  a  few 
times  in  the  life  of  the  bridge.  Others 
again  provide  for  these  extra  stresses  due 
to  wind  when  they  exceed  25  per  cent,  of 
the  stresses  due  to  the  dead  and  live-load 
combined,  by  specifying  that  the  section 
of  member  shall  be  increased  until  the 
total  stress  (including  wind)  per  square 


§67]  174 

inch  shall  not  exceed  by  more  than  25  per 
cent,  the  maximum  fixed  for  live  and  dead 
load  only.  The  members  C  E  and  E  G 
are  the  only  ones  of  this  truss  that  fall  un- 
der the  last  rule. 

The  stresses  in  the  end  (batter)  posts  are 
increased  over  those  given  above  by  the 
action  of  the  wind  transferred  to  the  tops 
of  the  end  posts  by  the  upper  lateral  brac- 
ing (Art.  63),  which,  through  the  medium 
of  the  portal  bracing,  causes  additional 
stresses  in  the  end  posts  from  the  bending 
moments  induced.  Lack  of  space  prevents 
a  discussion  of  portal  bracing  here.  Cer- 
tain forms  of  portal  bracing  were  discussed 
by  the  author  in  Engineering  News  for 
July  13,  1889,  but  a  very  full  analysis  will 
be  found  in  Johnson's  Framed  Structures, 
to  which  the  reader  is  referred. 

67.  Determination  of  Theoretical  Sec- 
tional Areas. — For  finding  the  sectional 
areas  of  the  members  of  the  vertical  truss 
the  following  formulas*  for  unit  stress  in 

*  See  Transactions  Am.  Soc.  C.  E.  for  June,  1886,  and  for 
Feb.  1892,  for  discussions  on  formulas  for  unit  stresses. 
Specifications  differ  very  much  with  regard  to  unit  stresses. 


175 


[§67 


thousand  pounds  per  square  inch  for  steel 
will  be  used  : 

For  ties,  a  =  10  (1  +  6) 


In  these  formulas, 

Q        Minimum  stress  on  member 
Maximum  stress  on  member 
I  =  length  of  member  in  inches 
r  =   least  radius  of   gyration  of  mem- 
ber in  inches. 


Member 

M:n. 

Stress 

Max. 
Stress 

6 

a 

J 

r 

Unit 
Stress 

Area 

ACE 

26.0 

129.4 

.20 

12.0 

12.0 

10.8 

EG 

41.5 

199.3 

.21 

12.1 

12.1 

16.5 

BE 

17.1 

137.7 

.123 

10.1 

10.1 

13.6 

DG 

0 

72.6 

0 

10.0 

10.0 

7.3 

EF 

0 

23.2 

0 

10.0 

10.0 

2.3 

BC 

9.0 

74.6 

.12 

11.2 

11.2 

6.7 

D  E 

4.5 

62.1 

.07 

10.1 

90 

6.0 

10.4 

FG 

4.5 

22.9 

.20 

12.0 

125 

5.2 

4.4 

AB 

42.9 

212.8 

.20 

12.0 

70 

8.2 

25.9 

BD 

41.5 

199.3 

.208 

12.1 

43 

9.7 

20.5 

DF 

46,7 

223.5 

.209 

12.1 

43 

9.7 

23.1 

The    maximum   and    minimum    stresses 
are  those  due  to  live  and  dead  loads  only, 


§67]  176 

and  are  taken  from  Art.  57.     The  values 

of  • —  are  assumed. 
r 

We  first  find  6  by  dividing  min.  stress  by 
max.  stress;  then  a  =  10  (1  +  6),  giving 
the  "unit  stress"  for  ties.  The  "unit 
stress  "  b  for  columns  is  then  found  for  the 
assumed  values  of  I  -~  r  by  the  formula 
above.  On  dividing  the  "max.  stress" 
by  "  unit  stress,"  the  "  area  "  of  the  sec- 
tion of  member  in  square  inches  is  found. 
Commercial  sizes  of  "  shapes "  must  be 
used.  This  will  alter  the  net  section  of  the 
ties  very  little;  of  the  posts  and  upper 
chords  more.  The  sections  of  the  latter, 
though,  are  often  very  much  altered  to  sat- 
isfy the  conditions  of  assembling  the  parts 
on  the  pins,  particularly  as  a  minimum 
thickness  of  metal  is  always  specified. 

In  fact,  it  may  be  well  to  conclude  with 
the  remark  that  the  work  has  only  fairly 
begun  when  the  stresses  in  a  bridge 
have  been  computed  and  "  designing  "  has 
been  reached,  of  which  only  the  simplest 
elements  have  been  given  above. 


177 

APPENDIX. 

MOST  ECONOMICAL,  HEIGHT  OF  TRUSSES  WITH 
PARALLEL  CHORDS. — The  weight  of  the  material 
of  a  truss  that  varies  with  its  height  h  is  a  function 
of  h  ;  denote  it  by  F  (A).  This  weight,  for  chords 
and  web,  can  be  found  by  multiplying  the  theo- 
retical section  (Art.  67)  of  each  member  by  its 
length  in  feet  multiplied  by  a  constant,  and  sum- 
ming the  products  for  the  entire  truss.  The  other 
parts  of  the  truss  (pins,  rivets,  cover  plates,  floor 
system,  transverse  bracing,  etc.)  do  not  vary 
perceptibly  with  h ;  hence  F  (7i)  is  the  weight  of 
material  computed  on  chord  and  web  stresses  only, 
and  it  is  a  minimum  when 

d¥(h)  F  (7*  +  A  h)  -  F  (7Q 

-^-=lme-          -rr-  =  *....(!) 

Denote  the  weight  of  both  chords  that  varies 
with  7i  by  Wc  ;  also  denote  the  variable  part  of 
the  weight  of  a  web  member  by  w,  and  call  its  in- 
clination to  the  vertical  i  and  its  length  I.  When 
the  height  of  the  truss  is  changed  to  h  -[-  A  h  (fig. 
29),  call  the  new  value  of  I,  I  +  A  I. 

The  formulas  for  unit  stresses  per  square  inch 
are  (see  Art.  67  for  meaning  of  6  and  r) : 


//r 
#X\ 

—  /r^~" 

/x\x 

•~w 
/i\v 

178 
Forties,  a  =  u  (1  -f  6)  ............................  (2) 

For  columns,  6  —  all  —  c  -J   .....................  (3) 

If  we  suppose  the  panel-length  and  unit  stresses 
in  the  chords  to  remain  invariable,  since  the  arms  of 
the  chord  members  are  li  and  h  -|-  A  h  before  and 
after  the  change  in  height,  the  new  stresses  in 
them  are  h  -+-  (h  +  A  7i)  of  the  stresses  before 
the  increase  in  height,  andhence  the  new  weight  of 
the  variable  material  in  the  chord  is 


If  S  denotes  the  shear  on  a  web  strut  the  old 

S  0&t          SZ2 

stress  is  S  sec  i  and  its  volume  —  ^  —  I  =  ~^~; 

S  (I  +  A  O2 
similarly  the  new  volume  is    ^  (h  -\-  &  ti  '        (  re 

b  is  given  by  e  #,  (3)  and  b  by  the  same  formula, 
when  I  is  changed  to  (I  -f-  A  I).  For  ties,  b  and 
b  are  replaced  by  a. 

The  new  weight  w'  of  the  variable  material  in  a 
web  member  is  thus  equal  to  w  multiplied  by  the 
ratio  of  new  to  old  volumes,  or, 
(/  +  A  0  3  h          b 

w'  =  w  ~  -~T^~      ~~^h  ~b  >  ................  r°' 

For  ties,  the  ratio  b  -r-  b  is  to  be  replaced  by  1. 
Actually  dividing  the  value  of  b  given  by  (3) 
by  the  value  of  J',  we  find,  regarding  r  as  constant, 


179 


*    =  J  +  A  J  -  X  "^ 


c^ 

r 
If  we  place  k  —  -  — - —  —  ,  this  ratio  can 

l->(*+A   I) 

b 
be  written,   ^  =  1  +  k  A  I (6) 

where  k  for  ties  =  0.     On  substituting  (6)  in  (5), 

(Z-LAZ)2        h                (l+&l)*hk<M 
w  =w ,-  '   


h+  A  A  J» 

P  —  Z8  A   h 


The  sum  of  the  new  weights  of  all  the  web 
members  will  be  designated  by  2  (V),  the  sum  of 
the  old  by  2  (M), 

Therefore,  as  F  (h)  =  Wc  -f  2  (w) 
F  (h  +  A  7i)  =  W'c  +  2  (w') 
F  (h  -I-  A  A)  -  F  (A)  _  Wc 

A  A  ~  ~  h  +  A 


from  eq.  (4) 


Hence,  writing  out  ^  («?')  —   2  (w),  dividing 

A  I 
by  A  h  and  noting  that  Urn  —  j-  =  c0*  »,  as  A  A 


180 

and  A  I  diminish  indefinitely,  we  have  for  a  mini- 
mum weight  of  truss  by  (1),  on  taking  limits  : 

—  Wc          ^  /Zhcosi—l  \ 

—  ^  —  -|-  2  (  --  ^  ---  -|-  COB  i  X  Mm  k  j  w  =  o. 

Since  cos  i  =  7i  -f-  1,  the  above  becomes  on  multiply- 
ing through  by  h, 

F2  h2  li*  "1 

We  =  Sr-p-  —l+jjl  (Urn  *)J  w 

2h  2 
Now,  ~rr~  —  1  —  2  cos2  *'  —  1  =:  cos  2  e,  and  plac- 

ing I  (lim  k)  =  m, 

I 

.'.m=   —  ~    .........  (8) 


we  have  finally, 

We  =  2w(cos2i  +  mcos2i)....  (9) 
where  m  is  given  by  (8)  for  struts  and  m  =  o  for 
ties.* 

ff,  therefore,  we  multiply  the  variable  weight  of 
each  web  member  by  (cos  2  i  +  #&  cos3  t)  C0rr^ 
spending  to  it,  the  sum  of  the  products  should  equal 
the  weight  of  the  two  chords,  if  the  most  economical 
height  has  been  chosen. 

For  vertical  members,  i  =  o,  cos  i  =  cos  2  i  =  1. 
When  t  <  45°,  cos  2  £  is  +  ;  i  —  45°,  cos  2  i  =  0; 
i  >  45°,  cos  2  i  is  minus. 

*  This  formula  being  true  for  a  Queen  Post  truss  is  like- 
wite  true  when  the  middle  panel  reduces  to  zero,  or  for  an 
A  truss  through  or  deck. 


181 

In  Van  Nostrand's  Magazine  for  January,  1877, 
Emu  Adler,  C.  E.,  deduced  the  formula  Wc  =  2 
(w  cos  2  i)  for  the  most  economical  depth  when  b 
is  constant. 

To  apply  (9),  since  the  weight  of  any  member 
is  equal  to  its  length  in  feet  X  cross-section  in 
inches  X  e,  where  e  is  the  same  for  all  members 
when  they  are  of  the  same  material,  make  out  the 
weight  of  each  member  regarding  e  as  unity  and 
substitute  numerical  results  for  the  web  members 
in  a  table  headed  as  follows  : 


Member 

w 

i 

cos  2  i  -\-mcos*  i 

Product 

The  product  w  (cos  2  i  -}-  m  cos2  i}  is  entered  in 
the  last  column,  and  the  sum  of  such  products  for 
all  the  web  members,  should  by  (9)  equal  "Wc,  the 
weight  of  the  two  chords,  if  the  most  economical 
height  has  been  chosen.  If  We  is  too  large,  then 
the  height  of  truss  may  be  increased,  and  vice 
versa. 

The  method  is  more  nearly  correct  for  long  spans, 
where  a  larger  number  of  members  proportionately 
have  sections  near  the  theoretical  sections.  Some 
interesting  conclusions  can  be  drawn  from  (9). 


182 

(a)  In  a  Warren  girder,  if  t  =  45°,  cos  2  i  =  o 
and  2"  (w  ?/i  cos  2  i)  is  quite  small,  so  that  for  (9) 
to  holJ,  Wc  must  be  small  or  the  height  of  the 
truss  large,  which  does  not  obtain  for  usual  panel- 
lengths.     It  it  thus  absurd  to  have  the  web  mem- 
bers inclined  45°  if  the  least  weight  of  truss  is  de- 
sired. 

(b)  In  whipple  or  other  multiple  systems,  cos  2 
i  is  nearer  zero  than  for  single  intersection  trusses 
of  the  same  number  of  panels,  hence  the  depth  of 
truss  is  greater  for  the  multiple  systems. 

(c)  As  a  rule,  angle  i  should  be  less  than  45°. 

(d)  The  greater  the  number  of  panels  for  the 
same  span,  the  heavier  the  web  for  the  same  kind 
of  truss ;  hence  by  (9),  We  must  be  greater  or  the 
truss  lower  for  economy. 

(e)  For   continuous  girders  it   is  known  that 
We  is  less  than  for  a  simple  girder  of  the  same 
span  having  the  same  weight  of  web ;  hence,  if  a 
certain  height  of  truss  satisfies  the  equality  (9) 
for  the  simple  girder,  a  less  height  is  required  for 
economy  for  the  continuous  girder. 

(/)  If  the  web  in  a  deck  bridge  is  heavier  than 
the  web  in  a  through  bridge  of  the  same  design, 
the  height  of  truss  must  be  less. 


THE   VAN   NOSTRAND   SCIENCE    SERIES 


No.  47.    LINKAGES:    THE  DIFFERENT  FORMS 

and  Uses  of  Articulated  Links.     By  J.  D.  C.  De  Roos. 

No.  48.    THEORY     OF     SOLID     AND     BRACED 

Elastic  Arches.  By  William  Cain,  C.E.  Second  edi- 
tion, revised  and  enlarged. 

No.  49.    MOTION  OF  A  SOLID  IN  A  FLUID.     By 

Thomas  Craig,  Ph.D. 

No.  50.    DWELLING-HOUSES;      THEIR     SANI- 

tary  Construction  and  Arrangements.  By  Prof.  W.  t^. 
Corfield. 

No.  51.    THE  TELESCOPE:    OPTICAL  PRINC5- 

pies  Involved  in  the  Construction  of  Refracting  and 
Reflecting  Telescopes,  with  a  new  chapter  on  the 
Evolution  of  the  Modern  Telescope,  and  a  Bibliography 
to  date.  With  diagfams  and  folding  plates.  By 
Thomas  Nolan.  Second  edition,  revised  and  enlarged. 

No.  52.      IMAGINARY     QUANTITIES;       THEIR 

Geometrical  Interpretation.  Translated  from  {he 
French  of  M.  Argand  by  Prof.  A.  S.  Hardy. 

No.  53.    INDUCTION  COILS;    HOW  MADE  AND 

How  Used.     Eleventh  American  edition. 

No.  54.    KINEMATICS     OF     MACHINERY.      By 

Prof.  Alex.  B.  W.  Kennedy.  With  an  Introduction  by 
Prof.  R.  H.  Thurston. 

No.  55.    SEWER  GASES;    THEIR  NATURE  AND 

Origin.  By  A.  de  Varona.  Second  edition,  revised  and 
enlarged. 

*No.  56.  THE  ACTUAL  LATERAL  PRESSURE 

of  Earthwork.     By  Benj.  Baker,  M.  Inst.,  C.E. 

No.  57.      INCANDESCENT    ELECTRIC    LIGHT- 

ing.  A  Practical  Description  of  the  Edison  System. 
By  L.  H.  Latimer.  To  which  is  added  the  Design  and 
Operation  of  the  Incandescent  Stations,  by  C.  J.  Field; 
and  the  Maximum  Efficiency  of  Incandescent  Lamps, 
by  John  W.  Howell. 

No.  58.     VENTILATION   OF   COAL  MINES.      By 

W.  Fairley,  M.E.,  and  Geo.  J.  Andr6. 

No.  59.    RAILROAD  ECONOMICS;    OR,  NOTES 

With  Comments.     By  S.  W.  Robinson,  C.E. 

No.  60.     STRENGTH        OF       WROUGHT-IRON 

Bridge  Members.     By  S.  W.  Robinson,  C.E. 
No.  61.    POTABLE  WATER,  AND  METHODS  OF 

Detecting  Impurities.     By  M.  N.  Baker.     Second  edi- 
tion, revised  and  enlarged. 
No.  63.     THEORY    OF    THE    GAS-ENGINE.      By 

Dougald  Clerk.  Third  edition.  With  additional 
matter.  Edited  by  F.  E.  Ideli,  M.E. 


THE  VAN   NOSTRAND   SCIENCE    SERIES 

No.  63.    HOUSE-DRAINAGE     AND     SANITARY 

Plumbing.     By  W.  P.  Gerhard.     Twelfth  edition. 

No.  64.    ELECTROMAGNETS.     By  A.  N.  Mans- 
field.    Second  edition,  revised. 

No.  65.    POCKET     LOGARITHMS     TO     FOUR 

Places  of  Decimals.  Including  Logarithms  of  Num- 
bers, etc. 

No.  66.    DYNAMO-ELECTRIC  MACHINERY.   By 

S.  P.  Thompson.  With  an  Introduction  by  F.  L.  Pope. 
Third  edition,  revised. 

No.  67.     HYDRAULIC   TABLES  FOR  THE   CAL- 

culation  of  the  Discharge  through  Sewers,  Pipes,  and 
Conduits.  Based  on  "Kutter's  Formula."  By  P.  J. 
Flynn. 

No.  68.     STEAM-HEATING.     By   Robert   Briggs. 

Third  edition,  revised,  with  additions  by  A.  R.  Wolff.  ^ 

No.  69.    CHEMICAL   PROBLEMS.      By  Prof.  J.  C. 

Foye.     Fifth  edition,  revised  and  enlarged. 

No.  70.    EXPLOSIVE       MATERIALS.     By    Lieut. 

John  P.  Wisser. 

No.  71.     DYNAMIC       ELECTRICITY.     By    John 

Hopkinson,  J.  N.  Shoolbred,  and  R.  E.  Day. 
No.  72.     TOPOGRAPHICAL       SURVEYING.      By 

George  J.  Specht,  Prof.  A.  S.  Hardy,  John  B.  McMaster, 

and  H.  F.  Walling.     Fourth  edition,  revised. 
No.  73.    SYMBOLIC  ALGEBRA;   OR,  THE  ALGE- 

bra  of  Algebraic  Numbers.     By  Prof.  William  Cain. 
No.  74.    TESTING     MACHINES;      THEIR    HIS- 

tory,  Construction  and  Use.     By  Arthur  V.  Abbott. 
No.  75.    RECENT     PROGRESS     IN     DYNAMO- 

electric  Machines.     Being  a  Supplement  to  "Dynamo- 
electric  Machinery.     By  Prof.  Sylvanus  P.  Thompson. 
No.  76.    MODERN    REPRODUCTIVE    GRAPHIC 

Processes.     By  Lieut.  James  S.  Pettit,  U.S.A. 
No.  77.    STADIA     SURVEYING.     The  Theory  of 

Stadia    Measurements.     By   Arthur   Winslow.     Eighth 

edition. 
No.  78.    THE      STEAM  -  ENGINE      INDICATOR 

and  Its  Use.     By  W.  B.  Le  Van. 
No.  79.     THE    FIGURE    OF    THE    EARTH.     By 

Frank  C.  Roberts,  C.E. 
No.  80.    HEALTHY          FOUNDATIONS          FOR 

Houses.     By  Glenn  Brown. 
*No.  81.   WATER     METERS:       COMPARATIVE 

Tests  of  Accuracy,  Delivery,  etc.     Distinctive  Features 

of    the    Worthington,    Kennedy,    Siemens,    and    Hesse 

meters.     By  Ross  E.  Browne. 


THE   VAN   NOSTRAND   SCIENCE    SERIES 


No.  82.    THE  PRESERVATION  OF  TIMBER  BY 

the  Use  of  Antiseptics.  By  Samuel  Ba'gster  Boulton, 
C.E. 

No.  83.    MECHANICAL,      INTEGRATORS.         By 

Prof.  Henry  S.  H.  Shaw,  C.E. 

No.  84.  FLOW    OF    WATER    IN    OPEN    CHAN- 

nels,  Pipes,  Conduits,  Sewers,  etc.  With  Tables.  'By 
P.  J.  Flynn,  C.E. 

No.  85.   THE      LUMINIFEROUS      AETHER.      By 

Prof.  De  Volson  Wood*. 

No.  86.    HANDBOOK    OF    MINERALOGY;     DE- 

termination,  Description,  and  Classification  of  Minerals 
Found  in  the  United  States.  By  Prof.  J.  C.  Foye. 
Fifth  edition,  revised. 

No.  87.    TREATISE  ON  THE  THEORY  OF  THE 

Construction  of  Helicoidal  Oblique  Arches.  By  John 
L.  Culley,  C.E. 

*No.  88.  BEAMS  AND  GIRDERS.  Practical  For- 
mulas for  their  Resistance.  By  P.  H.  Philbrick. 

No.  89.   MODERN  GUN  COTTON:    ITS  MANTJ- 

facturo,  Properties,  and  Analyses.     By  Lieut.  John  P. 

Wisser,  U.S.A. 
No.  90.  ROTARY    MOTION    AS    APPLIED     TO 

the  Gyroscope.     By  Major  J.  G.  Barnard. 
No.  91.  LEVELING:    BAROMETRIC,  TRIGONO- 

metric,    and    Spirit.     By    Prof.    I.    O.    Baker.     Third 

No.  92.  PETROLEUM;   ITS  PRODUCTION  AND 

Use.     By  Boverton  Redwood,  F.I.C.,  F.C.S. 

No.  93.  RECENT  PRACTICE  IN  THE  SANI- 
tary  Drainage  of  Buildings.  With  Memoranda  on  the 
Cost  of  Plumbing  Work.  Second  edition,  revised  and 
enlarged.  By  William  Paul  Gerhard,  C.E. 

No.  94.  THE  TREATMENT  OF  SEWAGE.  By 
Dr.  C.  Meymott  Tidy. 

No.  95.  PLATE-GIRDER  CONSTRUCTION.  By 
Isami  Hiroi,  C.E.  Fourth  edition,  revised. 

No.  96.  ALTERNATE-CURRENT  MACHINERY. 
By  Gisbet  Kapp,  Assoc.  M.  Inst.,  C.E. 

No.  97.  THE  DISPOSAL  OF  HOUSEHOLD 
Wastes.  Second  edition.  By  W.  Paul  Gerhard,  Sani- 
tary Engineer. 

No.  98.  PRACTICAL  DYNAMO-BUILDING  FOR 
Amateurs.  How  to  Wind  for  Any  Output.  By  Fred- 
erick Walker.  Fully  illustrated.  Third  edition. 

No.  99.  TRIPLE-EXPANSION  ENGINES  AND 
Engine  Trials.  By  Prof.  Osborne  Reynolds.  Edited 
with  notes,  etc.,  by  F.  E.  Idell,  M.E. 


THE   VAN   NOSTRAND   SCIENCE    SERIES 


No.  100.   HOW  TO  BECOME  AN  ENGINEER;   or, 

The  Theoretical  and  Practical  Training  necessary  in 
Fitting  for  the  Duties  of  the  Civil  Engineer.  By  Prof. 
Geo.  W.  Plympton. 

No.  101.  THE      SEXTANT,   and   Other  Reflecting 

Mathematical  Instruments.  With  Practical  Hints  for 
their  Adjustment  and  Use.  By  F.  R.  Brainard,  U.  S. 
Navy. 

No.  102.   THE     GALVANIC     CIRCUIT    INVESTI- 

ga^ed  Mathematically.  By  Dr.  G.  S.  Ohm,  Berlin, 
1827.  Translated  by  William  Francis.  With  Preface 
and  Notes  by  the  Editor,  Thomas  D.  Lockwood, 
M.I.E.E.  Second  edition. 

No.  103.  THE  MICROSCOPICAL  EXAMINATION 

of  Potable  Water.  With  Diagrams.  By  Geo.  W. 
Rafter.  Second  edition. 

No.  104.  VAN  NOSTRAND'S  TABLE-BOOK  FOR 

Civil  and  Mechanical  Engineers.  Compiled  by  Prof. 
Geo.  W.  Plympton. 

No.  105.     DETERMINANTS.     An  Introduction   to 

the  Study  of,  Wi'th  Examples  and  Applications.  By 
Prof.  G.  A.  Miller. 

No.  106.   COMPRESSED  AIR.     Experiments  upon 

the  Transmission  of  Power  by  Compressed  Air  in  Paris. 
(Popp's  System.)  By  Prof.  A.  B.  W.gKennedy.  The 
Transmission  and  Distribution  of  Power  from  Central 
Stations  by  Compressed  Air.  By  Prof.  W.  C.  Unwin. 
Edited  by  F.  E.  Idell.  Vhird  edition.  j 

No.  107.  A  GRAPHICAL  METHOD  FOR  SWING 

Bridges.  A  Rational  and  Easy  Graphical  Analysis  of 
the  Stresses  in  Ordinary  Swmg  Bridges.  With  an  Intro- 
duction on  the  General  Theory  of  Graphical  Statics,  ivith 
Folding  Plates.  Second  edition.  By  Benjamin  F.  La 
Rue. 

No.  108.   SLIDE-VALVE   DIAGRAMS.     A   French 

Method  for  Constructing  Slidt-valve  Diagrams.  B/ 
Lloyd  Bankson,  B.S.,  Assistant  Naval  Constructor, 
U.  S.  Navy.  8  Folding  Plates. 

No.  109.  THE    MEASUREMENT   OF   ELECTRIC 

Currents.  Electrical  Measuring  Instruments.  By 
James  Swinburne.  Meters  for  Electrical  Energy.  By 
C.  H.  Wordingham.  Edited,  with  P.veface,  by  T.  Corr- 
merford  Martin.  With  Folding  Plate  and  Numerous 
Illustrations. 

No.  110.  TRANSITION     CURVES.     A  Field-book 

for  Engineers,  Containing  Rules  and  Tables  for  Lay 
out    Transition    Curves.     By    Walter    G.    Fox, 
Second  edition. 


THE  VAN  NOSTRAND   SCIENCE    SERIES 

No.  111.  GAS-LIGHTING    AND    GAS-FITTING. 

Specifications  and  Rules  for  Gas-piping.  Notes  on  the 
Advantages  of  Gas  for  Cooking  and  Heating,  and 
Useful  Hints  to  Gas  Consumers.  Third  edition.  By 
Wm.  Paul  Gerhard,  C.E. 

No.  112.  A   PRIMER   ON   THE   CALCULUS.     By 

E.  Sherman  Gould,  M.  Am.  Soc.  C.E.     Third  edition, 
revised  and  enlarged. 

No.  113.  PHYSICAL  PROBLEMS  and  Their  So- 
lution. By  A.  Bourgougnon,  formerly  Assistant  at 
Bellevue  Hospital.  Second  edition.  ^ 

No.  114.  USE      OF      THE      SLIDE      RULE.     By 

F.  A.  Halsey,  of  the  "American  Machinist."     Fourth 
edition,  revised  and  enlarged. 

No.  115.  TRAVERSE  TABLE.  Showing  the  Dif- 
ference of  Latitude  and  Departure  for  Distances  Between 
1  and  100  and  for  Angles  to  Quarter  Degrees  Between  1 
Degree  and  90  Degrees.  (Reprinted  from  Scribner's 
Pocket  Table  Book.)  Third  edition. 

No.  116.  WORM  AND  SPIRAL  GEARING.  Re- 
printed from  "American  Machinist."  By  F.  A.  Halsey. 
Second  revised  and  enlarged  edition. 

No.  117.  PRACTICAL      HYDROSTATICS,      AND 

Hydrostatic  Formulas.  With  Numerous  Illustrative 
Figures  and  Numerical  Examples.  By  E.  Sherman 
Gould. 

No.  118.  TREATMENT    OF    SEPTIC    SEWAGE, 

with  Diagrams  and  Figures.  By  Geo.  W.  Rafter.  Second 
edition. 

No.  119.  LAY-OUT  OF  CORLISS  VALVE  GEARS. 

With  Folding  Plates  and  Diagrams.  By  Sanford  A. 
Moss,  M.S.,  Ph.D.  Reprinted  from  "The  American 
Machinist,"  with  revisions  and  additions.  Second 
edition. 

No.  120.  ART  OF  GENERATING  GEAR  TEETH. 

By  Howard  A.  Coombs.  With  Figures,  Diagrams  and 
Folding  Plates.  Reprinted  from  the  "American  Ma- 
chinist." 

No.  121.  ELEMENTS     OF     GAS     ENGINE     DE- 

sign.  Reprint  of  a  Set  of  Notes  accompanying  a  Course 
of  Lectures  delivered  at  Cornell  University  in  1902.  By 
Sanford  A.  Moss.  Illustrated. 

No.  122.  SHAFT    GOVERNORS.     By    W.    Trinks 

and  C.  Housum.     Illustrated. 
No.  123.  FURNACE  DRAFT;    ITS  PRODUCTION 

by  Mechanical  Methods.  A  Handy  Reference  Book, 
with  figures  and  tables.  By  William  Wallace  Christie. 
Illustrated.  Second  edition,  revised. 


THE   VAN   NOSTRAND  SCIENCE    SERIES 

No.  134.   "  SUMNEB'S  METHOD  "  FOB     FIND- 

ing  a  Ship's  Position.  Condensed  and  Improved.  Bv 
Rev..G.  M.  Searle,  Ph.D. 

No.  125.   TABLES  FOB  THE  DETEBMINATION 

of  Common  Rocks.  By  Oliver  Bowles,  M.A.,  Instructor 
of  Geology  and  Mineralogy,  University  of  Minnesota. 

No.  126.   PBINCIPLES  AND  DESIGN  OF  AEBO- 

Planes.  By  Herbert  Chatley,  B.Sc.,  Author  of  "The 
Problem  of  Flight,"  "Force  of  the  Wind,"  etc.  Second 
edition,  revised.  Illustrated. 

No.  127.   SUSPENSION  BBIDGES  AND   CANTI- 

levers;  their  Economic  Proportions  and  Limiting  Spans, 
Second  Edition,  revised  and  enlarged.  By  D  B 
Steinman,  C.E.,  Ph.D.,  Professor  of  Civil  Engineering. 
University  of  Idaho. 


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